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Exponents-1
Algebra of Exponents
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
an = a · a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
an = a · a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
Examples: 53 = 5 · 5 · 5 = 125,
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
an = a · a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
Examples: 53 = 5 · 5 · 5 = 125,(
43
)2
= 43· 4
3= 16
9
It is then clear that the law am+n = aman must hold when m and n are positive integers, because
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
an = a · a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
Examples: 53 = 5 · 5 · 5 = 125,(
43
)2
= 43· 4
3= 16
9
It is then clear that the law am+n = aman must hold when m and n are positive integers, because
am+n = a · a · . . . · a · a · a︸ ︷︷ ︸m+n copies of a
and
Exponents-1
Algebra of ExponentsMastery of the laws of exponents is essential to succeed in Calculus. We begin with the simplest case:
Integer ExponentsSuppose n is a positive integer. Then an is simply means n copies of a multiplied together, i.e.,
an = a · a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
Examples: 53 = 5 · 5 · 5 = 125,(
43
)2
= 43· 4
3= 16
9
It is then clear that the law am+n = aman must hold when m and n are positive integers, because
am+n = a · a · . . . · a · a · a︸ ︷︷ ︸m+n copies of a
and
aman = a · a · . . . · a · a · a︸ ︷︷ ︸m copies of a
×a · a · . . . · a · a · a︸ ︷︷ ︸n copies of a
mean exactly the same thing.
Exponents-2
Powers You Should Know
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
In addition, you also know four third powers, or cubes: 13 = 1, 23 = 8, 33 = 27, 43 = 64,
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
In addition, you also know four third powers, or cubes: 13 = 1, 23 = 8, 33 = 27, 43 = 64,
and three fourth powers: 14 = 1, 24 = 16, 34 = 81.
The higher powers of 2 should be known at least up to the tenth power:
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
In addition, you also know four third powers, or cubes: 13 = 1, 23 = 8, 33 = 27, 43 = 64,
and three fourth powers: 14 = 1, 24 = 16, 34 = 81.
The higher powers of 2 should be known at least up to the tenth power:
25 = 32, 26 = 64, 27 = 128, 28 = 256, 29 = 512, and 210 = 1024.
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
In addition, you also know four third powers, or cubes: 13 = 1, 23 = 8, 33 = 27, 43 = 64,
and three fourth powers: 14 = 1, 24 = 16, 34 = 81.
The higher powers of 2 should be known at least up to the tenth power:
25 = 32, 26 = 64, 27 = 128, 28 = 256, 29 = 512, and 210 = 1024.
The powers of 10 are of course really easy:
10n is just a 1 followed by n 0’s.
Exponents-2
Powers You Should KnowFrom your knowledge of the multiplication tables from 1 to 10 you automatically know ten second powers, or
squares:
12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, and 102 = 100.
In addition, you also know four third powers, or cubes: 13 = 1, 23 = 8, 33 = 27, 43 = 64,
and three fourth powers: 14 = 1, 24 = 16, 34 = 81.
The higher powers of 2 should be known at least up to the tenth power:
25 = 32, 26 = 64, 27 = 128, 28 = 256, 29 = 512, and 210 = 1024.
The powers of 10 are of course really easy:
10n is just a 1 followed by n 0’s.
Any number raised to the “zero-th” power is defined to be 1, i.e.,
a0 = 1
Exponents-3
Examples:
40 = 1, π0 = 1,(
14
)0
= 1
Exponents-3
Examples:
40 = 1, π0 = 1,(
14
)0
= 1
Note that the law am+n = aman is still true when either m or n, or both, equal 0.
Negative Integer Powers
Exponents-3
Examples:
40 = 1, π0 = 1,(
14
)0
= 1
Note that the law am+n = aman is still true when either m or n, or both, equal 0.
Negative Integer Powers
We define a−1 = 1a
Exponents-3
Examples:
40 = 1, π0 = 1,(
14
)0
= 1
Note that the law am+n = aman is still true when either m or n, or both, equal 0.
Negative Integer Powers
We define a−1 = 1a
and a−n = 1an
Exponents-3
Examples:
40 = 1, π0 = 1,(
14
)0
= 1
Note that the law am+n = aman is still true when either m or n, or both, equal 0.
Negative Integer Powers
We define a−1 = 1a
and a−n = 1an
so that the law am+n = aman is still true when either m or n or both are integers less than 0.
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
Roots
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8.
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e.,
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e., 813 = 2.
We must be very careful to use precise language.
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e., 813 = 2.
We must be very careful to use precise language. We say that 2 is
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e., 813 = 2.
We must be very careful to use precise language. We say that 2 is the fourth root of 16,
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e., 813 = 2.
We must be very careful to use precise language. We say that 2 is the fourth root of 16, and we say that −2
is
Exponents-4
In addition, we have another useful law for quotients of powers of the same number a:
am
an= am−n
which is true for all possible integers m and n, be they positive or negative.
RootsThe n-th root of a positive number a is that positive real number a
1n (also written n√a) which, when raised to
the n-th power, returns a, i.e.,
(a
1n
)n = a
In particular:(a
12
)2 = a,(a
13
)3 = a.
As an example, 813 is that positive number which, when raised to the third power, gives 8. Since 23 = 8, then 2
is the number we are looking for, i.e., 813 = 2.
We must be very careful to use precise language. We say that 2 is the fourth root of 16, and we say that −2
is a fourth root of 16.
Exponents-5
Fractional Powers
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e.,
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n,
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n,
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or “a to the m over n.”
Examples
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or “a to the m over n.”
Examples
2723 =
(27
13
)2 = 32 = 9
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or “a to the m over n.”
Examples
2723 =
(27
13
)2 = 32 = 9
532 =
(5
12
)3 = √5 · √5 · √5 = 5√
5
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or “a to the m over n.”
Examples
2723 =
(27
13
)2 = 32 = 9
532 =
(5
12
)3 = √5 · √5 · √5 = 5√
5
(1
81
) 34 =
( 1
81
) 14
3
=[
13
]3
= 127
Even with these fractional exponents, we still have the law:
Exponents-5
Fractional PowersSuppose a is any positive real number, and r is any rational number, i.e., r = m
n, the quotient of two integers
m and n, where n �= 0.
We define ar = amn =
(a
1n
)m,
and call it “a raised to the r -th power,” or “a to the m over n.”
Examples
2723 =
(27
13
)2 = 32 = 9
532 =
(5
12
)3 = √5 · √5 · √5 = 5√
5
(1
81
) 34 =
( 1
81
) 14
3
=[
13
]3
= 127
Even with these fractional exponents, we still have the law:
a−r =(
1a
)r= 1ar
Exponents-6
Examples
Exponents-6
Examples
27−23 =
Exponents-6
Examples
27−23 = 1
2723
=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples:
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s)
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r 8−
r3 =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r 8−
r3 =
(8
13
)−r =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r 8−
r3 =
(8
13
)−r = 2−r =
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r 8−
r3 =
(8
13
)−r = 2−r = 12r
Multiplication and Division of Bases
Exponents-6
Examples
27−23 = 1
2723
= 1(27
13
)2 =1
(3)2 =19
5−32 = 1
532
= 1(5
12
)3 =1(√5)3 =
1
5√
5= 1
5√
5
√5√5=√
525
The Laws of Exponent Algebra
Addition and Subtraction of Exponents
ar+s = aras and ar−s = ar
as= 1as−r
Examples: 51+s = 5 (5s) 5(
12−s
)= 5
12
5s=√
55s
Multiplication and Division of Exponents
ars = (ar )s = (as)r and ars =
(a
1s
)r = (ar ) 1s
Examples: 22r = (22)r = 4r 8−
r3 =
(8
13
)−r = 2−r = 12r
Multiplication and Division of Bases
Exponents-7
(ab)r = arbr
Exponents-7
(ab)r = arbr and(ab
)r = arbr
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s .
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4,
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) =
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers).
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer,
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a;
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
3.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
3. A major goal of calculus is to define ar for r any real number, not just for r a rational number.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
3. A major goal of calculus is to define ar for r any real number, not just for r a rational number. It is
then amazing but true that the rules remain valid.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
3. A major goal of calculus is to define ar for r any real number, not just for r a rational number. It is
then amazing but true that the rules remain valid.
4. The rule a−r =(
1a
)r = 1ar is quite useful in allowing us to eliminate negative exponents in fractions.
Exponents-7
(ab)r = arbr and(ab
)r = arbr
1. Memorizing these rules is not difficult if you keep in mind the integer exponent case. For example,
suppose you need to simplify(
3r+1)r−1
but cannot recall if the rule for (ar )s is ars or ar+s . Just check the
formulas in a simple case, say with a = 2, r = 1, and s = 2:
(ar )s = (21)2 = 22 = 4, ars = 21·2 = 22 = 4, which are equal, but ar+s = 21+2 = 23 = 8.
This should pretty quickly help you remember (ar )s = ars . Thus(3r+1
)r−1 = 3(r+1)(r−1) = 3r2−1.
2. Although ar has only been defined for a a positive real number, in some cases ar does make sense when
a is negative. For example, (−2)3 = −8, and thus (−8)13 = −2. On the other hand, (−8)
12 is not defined (unless
we use complex numbers). In general, amn will be defined for negative a if n is an odd integer, but not if n is an
even integer. For these reasons the rules are not always valid for negative bases and must be handled with care!
As an example, when a is negative (say a = −2) it is NOT true that(a2) 1
2 = a; instead(a2) 1
2 = −a.
3. A major goal of calculus is to define ar for r any real number, not just for r a rational number. It is
then amazing but true that the rules remain valid.
4. The rule a−r =(
1a
)r = 1ar is quite useful in allowing us to eliminate negative exponents in fractions.
Observe its use in the following simplification:
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2=
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3=
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3=
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 =
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 = 4
12a
12 =
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 = 4
12a
12 = 2
√a,
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 = 4
12a
12 = 2
√a,
(a2
8
) 23
=
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 = 4
12a
12 = 2
√a,
(a2
8
) 23
=(a2) 2
3
823
=
Exponents-8
(x + 2)−3(x − 2)3
(x2 − 4)−2= (x − 2)3(x2 − 4)2
(x + 2)3
where the terms with negative exponents have “switched” between the numerator and denominator:
= (x − 2)3 [(x − 2)(x + 2)]2
(x + 2)3= (x − 2)3(x − 2)2(x + 2)2
(x + 2)3
= (x − 2)5(x + 2)2
(x + 2)3= (x − 2)5
x + 2
Examples: (4a)12 = 4
12a
12 = 2
√a,
(a2
8
) 23
=(a2) 2
3
823
= a43
4
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer.
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the n-th root of a.
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the n-th root of a. (when n = 2 we simply write√a in place of 2
√a.)
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the n-th root of a. (when n = 2 we simply write√a in place of 2
√a.)
From its definition we have the basic formulas:
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the n-th root of a. (when n = 2 we simply write√a in place of 2
√a.)
From its definition we have the basic formulas:( n√a)n = n√an = a
for any positive real number a.
We also haven√ab = n√a n
√b and
Exponents-9
Non-Formulas for the Addition and Subtraction of Bases
(a+ b)r =? and (a− b)r =?
There are NO simple formulas in these cases.
For instance, (a+ b)r does NOT equal ar + br .
This is a very common and fatal error, so be very, very careful!!
n-th RootsAn important special case of exponents occurs when r = 1
nfor n a positive integer. In that case we often use
the notation
a1n = n√a
and refer to n√a as the n-th root of a. (when n = 2 we simply write√a in place of 2
√a.)
From its definition we have the basic formulas:( n√a)n = n√an = a
for any positive real number a.
We also haven√ab = n√a n
√b and n
√ab=
n√an√b
Exponents-10
Exponents-10
Again, we emphasize that there are
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1),
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)=
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)= x(x + 1)
15 − (x + 1)(x + 1)
15
(x + 1)(x − 1)=
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)= x(x + 1)
15 − (x + 1)(x + 1)
15
(x + 1)(x − 1)=
(x + 1)15 [x − (x + 1)]
(x + 1)(x − 1)=
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)= x(x + 1)
15 − (x + 1)(x + 1)
15
(x + 1)(x − 1)=
(x + 1)15 [x − (x + 1)]
(x + 1)(x − 1)= (−1)(x + 1)−
15 (x + 1)1(x − 1)
=
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)= x(x + 1)
15 − (x + 1)(x + 1)
15
(x + 1)(x − 1)=
(x + 1)15 [x − (x + 1)]
(x + 1)(x − 1)= (−1)(x + 1)−
15 (x + 1)1(x − 1)
=
Exponents-10
Again, we emphasize that there are NO simple formulas forn√a+ b or
n√a− b
Don’t dream anything up for these expressions!!
Examples
Example 1 Simplifyx(x + 1)
15 − (x + 1)
65
x2 − 1
Solution: Using65= 1+ 1
5, and x2 − 1 = (x + 1)(x − 1), we rewrite the expression as
x(x + 1)15 − (x + 1)1+ 1
5
(x + 1)(x − 1)= x(x + 1)
15 − (x + 1)(x + 1)
15
(x + 1)(x − 1)=
(x + 1)15 [x − (x + 1)]
(x + 1)(x − 1)= (−1)(x + 1)−
15 (x + 1)1(x − 1)
= − 1
(x + 1)45 (x − 1)
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers.
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator,
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1= (x − 2)−
15 [1+ (x − 2)]x − 1
=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1= (x − 2)−
15 [1+ (x − 2)]x − 1
=
(x − 2)−15 (x − 1)
x − 1=
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1= (x − 2)−
15 [1+ (x − 2)]x − 1
=
(x − 2)−15 (x − 1)
x − 1= (x − 2)−
15 =
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1= (x − 2)−
15 [1+ (x − 2)]x − 1
=
(x − 2)−15 (x − 1)
x − 1= (x − 2)−
15 =
Exponents-11
Example 2 Simplify(x − 2)−
15 + (x − 2)
45
x − 1
Solution: This is a fairly common type of expression. As in the previous example we have the same
base raised to different fractional powers. In general the way to proceed is to
factor out the lowest fractional power —in this case (x − 2)−15 — from both terms in the numerator, by using
the fact that (x − 2)−15 (x − 2)
15 = 1:
(x − 2)−15 + (x − 2)
45
x − 1= (x − 2)−
15 + 1 · (x − 2)
45
x − 1=
(x − 2)−15 +
[(x − 2)−
15 (x − 2)
15
]· (x − 2)
45
x − 1=
(x − 2)−15 + (x − 2)−
15
[(x − 2)
15 · (x − 2)
45
]x − 1
=
(x − 2)−15 + (x − 2)−
15 (x − 2)
15+ 4
5
x − 1= (x − 2)−
15 + (x − 2)−
15 (x − 2)
x − 1= (x − 2)−
15 [1+ (x − 2)]x − 1
=
(x − 2)−15 (x − 1)
x − 1= (x − 2)−
15 = 1
5√x − 2
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
(1)(√c −
√d)(√
c +√d)=
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
(1)(√c −
√d)(√
c +√d)= (√c)2 −
(√d)2 =
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
(1)(√c −
√d)(√
c +√d)= (√c)2 −
(√d)2 = c − d
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
(1)(√c −
√d)(√
c +√d)= (√c)2 −
(√d)2 = c − d
(2)(a−
√b)(a+
√b)= a2 −
(√b)2 = a2 − b
Exponents-12
Example 3 Simplify 3√
8a6(x − h)4 + 2a2h 3√x − h
Solution: The trick in this type of an expression is to “pull” as much out of the cube root as possible.
We start by examining the first term:
3√
8a6(x − h)4 = 3√
83√a6 3√(x − h)4 = 3
√23 3√(a2)3 3
√(x − h)3(x − h) =
2a2(x − h) 3√x − h
Thus our full expression becomes
2a2(x − h) 3√x − h+ 2a2h 3
√x − h = 2a2 3
√x − h(x − h+ h) = 2a2x 3
√x − h
Algebraic Conjugates
The terms (1)√c −√d and
√c +√d are called algebraic conjugates of each other, as are pairs of terms like
(2) a−√b and a+√b and (3)√a− b and
√a+ b.
The difference of squares law gives us their products:
(1)(√c −
√d)(√
c +√d)= (√c)2 −
(√d)2 = c − d
(2)(a−
√b)(a+
√b)= a2 −
(√b)2 = a2 − b
(3)(√a− b) (√a+ b) = (√a)2 − b2 = a− b2
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√x + 1−√x
(√x + 1+√x√x + 1+√x
)=
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√x + 1−√x
(√x + 1+√x√x + 1+√x
)=
√x + 1+√x
(√x + 1−√x)(√x + 1+√x) =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√x + 1−√x
(√x + 1+√x√x + 1+√x
)=
√x + 1+√x
(√x + 1−√x)(√x + 1+√x) =
√x + 1+√xx + 1− x =
Exponents-13
Example 4 Rewrite1√
c +√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c +√d
(√c −√d√c −√d
)=
√c −√d
(√c +√d)(√c −√d) =
√c −√dc − d .
Example 5 Rewrite1√
c −√d as a fraction with no roots in the denominator.
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√c −√d
(√c +√d√c +√d
)=
√c +√d
(√c −√d)(√c +√d) =
√c +√dc − d .
Example 6 Simplify1√
x + 1−√x .
Solution: Multiply by 1 in the form of the conjugate of the denominator over itself:
1√x + 1−√x
(√x + 1+√x√x + 1+√x
)=
√x + 1+√x
(√x + 1−√x)(√x + 1+√x) =
√x + 1+√xx + 1− x =
Exponents-14
√x + 1+√x
1=
Exponents-14
√x + 1+√x
1=
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√x
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer:
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1.
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1. Then√a2 =
√(−1)2 = √1 = 1 �= −1
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1. Then√a2 =
√(−1)2 = √1 = 1 �= −1
√a2 is called the
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1. Then√a2 =
√(−1)2 = √1 = 1 �= −1
√a2 is called the absolute value of a and is never negative.
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1. Then√a2 =
√(−1)2 = √1 = 1 �= −1
√a2 is called the absolute value of a and is never negative. It is written |a|.
Exponents-14
√x + 1+√x
1= √
x + 1+√x .
Example 7 Remove all square roots from the denominator ofh√
x + h−√xSolution: We must rationalize the denominator.
h√x + h−√x =
h√x + h−√x ·
(√x + h+√x√x + h+√x
)=
h(√x + h+√x
)(√x + h−√x
)(√x + h+√x
) =h(√x + h+√x
)(x + h− x) =
h(√x + h+√x
)h
=√x + h+√x
Example 8
Is it always true that√a2 = a?
Answer: NO . Let a = −1. Then√a2 =
√(−1)2 = √1 = 1 �= −1
√a2 is called the absolute value of a and is never negative. It is written |a|.
Exponents-15
Many calculus errors are made when an expression of the form√a2 is incorrectly replaced by a.
Example 9 Determine all x values for which√(x − 1)2 = 1.
Exponents-15
Many calculus errors are made when an expression of the form√a2 is incorrectly replaced by a.
Example 9 Determine all x values for which√(x − 1)2 = 1.
Solution: We know our equation is true if (x − 1)2 = 1.
Exponents-15
Many calculus errors are made when an expression of the form√a2 is incorrectly replaced by a.
Example 9 Determine all x values for which√(x − 1)2 = 1.
Solution: We know our equation is true if (x − 1)2 = 1. Since the only numbers whose square is 1 are
−1 and 1,
Exponents-15
Many calculus errors are made when an expression of the form√a2 is incorrectly replaced by a.
Example 9 Determine all x values for which√(x − 1)2 = 1.
Solution: We know our equation is true if (x − 1)2 = 1. Since the only numbers whose square is 1 are
−1 and 1, we either have x − 1 = −1 or x − 1 = 1
Exponents-15
Many calculus errors are made when an expression of the form√a2 is incorrectly replaced by a.
Example 9 Determine all x values for which√(x − 1)2 = 1.
Solution: We know our equation is true if (x − 1)2 = 1. Since the only numbers whose square is 1 are
−1 and 1, we either have x − 1 = −1 or x − 1 = 1 so either x = 0 or x = 2.
