Done by: Wa’ad Qayed Al-Johara Al-Thani Moza Salam Zahra Abdulla For Dr.Fuad Almuhanadi

Done by:

Wa’ad QayedAl-Johara Al-Thani

Moza SalamZahra Abdulla

ForDr.Fuad Almuhanadi

1. Exponents: Basic Rules (zahra+Wa’ad)

2. Exponential Equations. (Moza)

3. Exponential function and their graphs.

(Moza+ ALjohara)

1. Application on Exponential function. (Wa’ad)

Section: 1

Exponents are shorthand for multiplication process. The "exponent" stands for how many times the thing is being multiplied. For ex. (5)(5)(5)= 125 this the same as 53 .

53The thing that's being multiplied

is called the "base"

the exponent is the "power"

Basic rules 1- Simplify (x3)(x4):

)xm) ( xn = (x(m+n)

So, (x3)( x4) = (x3+4 ) =x7

However we can’t simplify

)x4 )(y3 =(xxxxyyy = (x4)(y3).

)x3 )(x4) = (xxx)(xxxx(

=xxxxxxx

=x7

2-Simplify (x2) 4

)xm( n= xmn

=  xxxxxxxx

=  x8

)x2 (4= x(2.4)= x8

)x2(4 = (x2)(x2)(x2)(x2)

) =xx)(xx)(xx)(xx(

So,

Solve (xy2)3

1- (xy 2) 3 = (xy 2)(xy 2)(xy 2)

= (xxx)(y 2 y 2 y 2)

= (xxx)(yyyyyy) = x 3y 6 = (x) 3 (y 2) 3

• This will only applied for multiplication and division. However we cannot apply it for addition and subtraction.

2

22

y

x

y

x

2-

For example (3+2)2

If we distribute the power 2 then,

(3 ) 2 +(2) 2 = 9+4 =13

And if we not distribute the power 2 then,

(3+2) 2 = (5) 2 = 25 = 13

Example (x – 2) 2 (x-2)2 = (x – 2) (x – 2)

= x2– 4x + 4

=xx – 2x – 2x + 4

35 = 36÷ 3 = 243 34 = 35 ÷ 3 = 81 33 = 34 ÷ 3 = 27 32 = 33 ÷ 3 = 9 31 = 32 ÷ 3 = 3 Then logically 30 = 31 ÷ 3 = 3 ÷ 3 = 1.

Anything to the power zero is just "1". (3)0= 1 a0 =1

m0 = m(n-n) = mn × m-n = mn ÷ mn = 1

3)[ -3x4y7z12(5 (–5x9y3z4)2]0 =1

negative exponent

• A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side.

• Write x-4 using only positive exponents.

• Write x2 / x-3 using only positive exponents. 5

32

3

2

3

2

1

1

1

1x

xx

x

x

x

x

4

44 1

1

1

x

xx

• Write 2x-1using only positive exponents

• Note that the "2" above does not move with the variable, the exponent is only on the "x".

xx

xx

22

1

22

1

11

• Write (3x)-2 using only positive exponents.

• Write (x -2/ y -3) -2 using only positive exponents

6

4

23

222

3

2

y

x

y

x

y

x

22

22

9

1

)3(

1

1

)3()3(

xx

xx

The radicalThe radical of any number can express in exponent.

or

2/122 244 2/1

288 3/13

38181 4/14

33333

222222

14/11/44/144 4

13/11/33/11/33/133 3

1- Solve 5x = 53.

x=3

2-Solve 32x–1 = 27.

32x–1 = 27 32x–1 = 33 2x – 1 = 3 2x = 4 x = 2

3- Solve 3x2–3x = 81.

3x2–3x = 81 3x2–3x = 34 x2 – 3x = 4 x2 – 3x – 4 = 0

(x – 4)(x + 1) = 0 x = –1, 4

4- Solve 42x2+2x = 8.

4 = 22

8 = 23

42x2+2x = (22)2x2+2x = 24x2+4x

Now I can solve:

42x2+2x = 8

24x2+4x= 23

4x2 + 4x = 3

4x2 + 4x – 3 = 0

(2x – 1)(2x + 3) = 0

x = 1/2 , –3/2

5- Solve 4x+1 = 1/64.

4x+1 = 1/64 4x+1 = 4–3

x + 1 = –3 x = –4

The general formula of the exponential function is:

F(x) = ax

• let a (01) (1) where a is the base such that a > 0 and a not equal

to 1.

As an example will take a=2

so, the function will be f(x) = 2x

To find the point of the function we will substitute in (x) with different

points.

xF(x)

-21/4

-11/2

01

12

24

f(x) = 2x

1

y

x

• As x increases with no bound, f(x) increases with no bound

Lim f (x) = positive infinity when (x) increase to positive infinity.

• As x decreases with no bound , f(x) approaches 0

Lim g (x) = 0 when (x) decrease to negative infinity.

• This means that the line y=0 works as horizontal asymptote.

• the value of the graph at zero will equal one (f (0) =2(0)=1).

• The domain of this graph is equal to the real numbers (R)

• the range of this graph is (0,)

g(x) = 3x

xg(x)

-21/9

-11/3

01

13

29

y

x

1

Comparison between f(x) and g(x)

1

y

x

F(x)=2x

G(x)=3x

• The domain of the function f(x)=ax is the real numbers (R).

• The range of the function f(x)=ax is the positive numbers between (0, )

• The greater is a the more rapidly the curve of the function increases (the more steeply) on the right of the y- axis and the faster it approaches the x-axis on the left of the y-axis. Therefore, the graph stretches more among the x-axis.

In the case of a (01) (1)

• Anther example of exponential function:

d(x) =1/2(2) x and g(x) = 2(2)x

the y-axis of the function g(x) = 2(2 )x multiplies by 2 and the function d(x) = 1/2(2 )x multiplies by half(1/2) . This appears when we find g(0)=2 and f(0)=1 we can see that it is multiply by 2 and the function stretch among the y-axis. And d(0)=1/2 and f(0)=1, the function shrink by 1/2

among the y-axis.

2

1

x

y

D(x)=1/2(2)x

G(x)=2(2)x

F(x)=(2)x

1 2 3 4-1-2-3-4

x1

2

3

4

-1

-2

-3

-4

y

g(x) = -(2 )^x

f(x) = (2 )^x

If we multiply by a negative number:

• range will be (0,-) and the domain equal the real number

• As x increases with no bound, g(x) decreases with no bound

Lim g (x) = positive infinity when (x) decrease to negative infinity.

• As x decreases with no bound , g(x) approaches 0

Lim g (x) = 0 when (x) decrease to negative infinity.

• This means that the line y=0 works as horizontal

asymptote.

• The third example:

g(x) = 2(2 )x+3 and d(x) = 1/2(2 )x-3

g(x) = 2(2 ) x

+3

d(x) = 2(2 ) x -3

• As x increases with no bound, g(x) increases with no bound

• Lim g (x) = positive infinity when (x) increase to positive infinity.

• As x decreases with no bound , g(x) approaches 3

• Lim g (x) = 3 when (x) decrease to negative infinity.

• This means that the line y=3 works as horizontal asymptote.

• The domain is the real number, but the range is equal to (3, )

g(x) = 2(2 )x +3

• As x increases with no bound, d(x) increases with no bound

• Lim g (x) = positive infinity when (x) increase to positive infinity.

• As x decreases with no bound , d(x) approaches -3

• Lim d (x) = -3 when (x) decrease to negative infinity.

• This means that the line y=-3 works as horizontal asymptote.

• The domain is the real number, but the range is equal to (-3, )

d(x) = 2(2 )x-3

• The fourth example:g(x)=2(-x)

g(x)=2(-

x)

f(x)=2x

•That the value of the graph at zero will equal one (g (0) =2 0=1). The domain of this graph is equal to the real numbers (R), and the range of this graph is (0,). •As x increases with no bound, f(x) approaches 0That is limf(x) =0, when (x) increases to positive infinity.•As x decreases with no bound, f(x) increases with no bound•This means that the line y=0 works as horizontal

1

x

y

• Example:

let a (01)

f(x) =1/2x xf(x)

-24

-12

01

11/2

21/4

1

y

x

• g(x) = 1/3x

xg(x)

-29

-13

01

11/3

21/9

y

x

1

•As x increases with no bound, g(x) approaches 0 That is limg(x) =0, when (x) increases to positive infinity.•As x decreases with no bound, g(x) increases with no boundThat is limg(x) =+∞ , when (x) decreases to negative infinity. •This means that the line y=0 works as horizontal •The greater is a the more rapidly the curve of the function increases ( the more steeply ). The more the graph stretch among the y-axis.

y

1x

G(x)=(1/3)x

F(x)=(1/2)x

If we added or subtracted from (X). To compare this is the example: function g(x)

= 2(2) ^(x+3),f(x) = 2(2) ^x and d(x) = 2(2) ^(x-3)

1 2 3 4-1-2-3-4

x1

2

3

4

-1

-2

-3

-4

yg(x) =

2(2)^(x+3)

d(x) = 2(2)^(x-3)

f(x) = 2(2)^(x)

On this application we will study the behavior of the antibiotic drugs' on the

body every period of time.The behavior of the drugs on the body

is the same as the behavior of the exponential function. The following

graph shows the behavior of the drug during the first 10 hours.

Time(t)amount of drug

010

0.59

18.3

1.57.8

27.2

2.56.7

36

3.55.3

45

4.54.5

54.3

5.54

63.7

6.53.3

73.1

7.52.7

82.5

8.52.1

91.9

9.51.8

101.5

amont of drug

0

2

4

6

8

10

12

0 5 10 15

Time

Am

ount

of d

rug

On the pervious graph we see that the amount of the drug among ten hours decline to 1.5. we use this graph to find

the equation of the line.

We use the general formula:

We chose the point (0, 10) to find first the value of a by substation on the pervious equation to be:

be

bte

010 be

10

tyb

0.59.0-0.211

1.08.3-0.186

1.57.8-0.166

2.07.2-0.164

2.56.7-0.160

3.06.0-0.170

3.55.3-0.181

4.05.0-0.173

4.54.5-0.177

5.04.3-0.169

5.54.0-0.167

6.03.7-0.166

6.53.3-0.171

7.03.1-0.167

7.52.7-0.175

8.02.5-0.173

8.52.1-0.184

9.01.9-0.185

9.51.8-0.181

10.01.5-0.190

total-3.51

average-0.176

By this equation we find the value of a to be 10 and it will be constant for this values. Now to find the value of be we will substitute on the equation to find the value and then find the

average.

By this the equation of the graph will be:

te 176.010

We apply the function to the amount of the drug on the blood for one day for each six hour

ty1y2 y3y4

010   

18.4   

27   

35.9   

44.9   

54.1   

63.513.5  

7 11.3  

8 9.5  

9 8  

10 6.7  

11 5.6  

12 4.714.7 

13  12.3 

14  10.3 

15  8.7 

16  7.3 

17  6.1 

18  5.115.1

19   12.7

20   10.6

21   8.9

22   7.5

23   6.3

24   5.3

Amount of the drugs for 24 hours

02468

10121416

0 5 10 15 20 25 30

time

Am

ou

nt

of

the

dru

gs

This is the values of the drugs for two days

Amount of drugs for48 hours

02468

1012141618

0 20 40 60

Time

Am

ou

nt

of

dru

gs

Amount of drugs for 96 hours

02468

1012141618

0 50 100 150

Time

Am

on

t o

f d

rug

sThis is the values of the drugs for four days

Observation:We can see that the value of the drugs on the bloodstream on the body remains constant and it will continue its maintaining and the patient stops taking the drug. So that way the doctors always told us to take the antibiotics after specific

time.