Technology: A Portal to Exploration and Discovery GCTM October 18th, 2012

Technology:A Portal to Exploration

and DiscoveryGCTM

October 18th, 2012Kenn Pendleton

[email protected]

Background:Student experiences prior to

this exploration

Using natural number exponents, the three laws of exponents were developed through exploration. This likely was accomplished using a scientific calculator.


this explorationAn exponent of zero was explored. A conjecture of the definition of a zero exponent was made and confirmed by applying the second law of exponents. The fact that the laws of exponents still held when exponents of zero were used was verified.


this explorationNegative exponents were explored. A conjecture of the definition of a negative exponent was made and confirmed by applying the second law of exponents. The fact that the laws of exponents still held when negative exponents were used was verified.

Part I:Exploring Fractional Exponents

tan

→ OFF AC /ON

Before turning on the calculator, notice the following keys.

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼

SHIFT OPTN VARS MENU

ALPHA x2 ^

EXITX,θ,T

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

REPLAY, or cursor: repeats processes and enables movement around the screen

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

MENU: accesses the main menu screen

OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

EXIT: returns to the previous menu level when nested menus are accessed.

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

Function keys: immediate access to screen functions and, when graphing, graph options. For example,

F1 F2 F3 F4 F5 F6

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,TAC/ON/OFF key: clears screens, turns the calculator on (and off, after having pressed SHIFT).

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

At the bottom-right, EXE(cute): performs intended operations and stores input EXE

tan

→ OFF AC /ON

F1 F2 F3 F4 F5 F6

▲

◄ REPLAY►

▼


ALPHA x2 ^

EXITX,θ,T

Turn on the calculator.

This is the Main Menu Screen.

We will start with Statistics activities. Notice the “2” in the upper-right corner.

Either enter “2” from the keyboard,

or cursor to the Statistics Icon and press EXE (cute) to select.

F1 F2 F3 F4 F5 F6

LIST 1 contains the exponents in the following expressions; LIST2 contains the values of the expressions.

4241404-14-2

F1 F2 F3 F4 F5 F6

Create a graph.

F1 F2 F3 F4 F5 F6

Your calculator is set to use GRAPH1.

F1 F2 F3 F4 F5 F6

Return to the main MENU .

Graph a function.

Enter “5,” or cursor to Graph and EXE .

A number of functions have been entered.Try to draw a graph.

F1 F2 F3 F4 F5 F6

EXIT back to the previous screen.

EXIT is below MENU .

F1 F2 F3 F4 F5 F6

With Y1 highlighted, SELECT this function.

F1 F2 F3 F4 F5 F6

DRAW the graph..

Notice the verb is highlighted.

F1 F2 F3 F4 F5 F6

This the same view window used earlier.

F1 F2 F3 F4 F5 F6

What is the

value of 4½ ?

Trace the function to find y when x = ½.

F1 F2 F3 F4 F5 F6

Trace

y = 4x; 4½ = 2What connection might students make between the expression and the result?

4½ = 4 × ½ = 2If the above conjecture were correct, what would be the value of the following:

8½ = ? 8 × ½ = 4

To test this conjecture, graph the function y = 8x and trace the graph to find the value of the function when x = ½ .

EXIT back to the Graph window.

Un-SELECT Y1; cursor to and SELECT Y2.

F1 F2 F3 F4 F5 F6

F1 F2 F3 F4 F5 F6

DRAW the graph of Y2.

Trace the graph to see whether 8½ = 4 as expected.

F1 F2 F3 F4 F5 F6

Trace

The conjecture was incorrect.

4½ = 2How else are 4 and 2 related?

How is that relation connected to the fractional exponent?

If the above conjecture were correct, what would be the value of 8½ ?

F1 F2 F3 F4 F5 F6

Return to the main MENU .

Arithmetic is done in Run-Matrix.

Enter “1,” or cursor to the icon and EXE .

The square root symbol is in gold above the x2 key (SHIFT: x2).

Find the square root of 8.

To change from simplest radical (F)orm to (D)ecimal form: F↔D (above “8”).

Return to the Graph window (MENU; 5).

This new conjecture proved to be true.

Un-SELECT Y2; SELECT Y3.

F1 F2 F3 F4 F5 F6



F1 F2 F3 F4 F5 F6

What is 9½ expected to be? Trace the graph.

F1 F2 F3 F4 F5 F6

Trace

2

F1 F2 F3 F4 F5 F6


Un-SELECT Y3; SELECT Y2; DRAW.

F1 F2 F3 F4 F5 F6

Trace

Trace the graph to x = 1/3 (0.3333333333).

Change the viewing window (V-Window).

F1 F2 F3 F4 F5 F6

V-Window

V-MEM; RECALL; 2; EXE .

F1 F2 F3 F4 F5 F6

EXIT back to Graph window.


F1 F2 F3 F4 F5 F6

Trace the graph to x = 1/3 (0.3333333333).

F1 F2 F3 F4 F5 F6

Trace

The value of the function is 2.

F1 F2 F3 F4 F5 F6


Un-SELECT Y2; SELECT Y1; DRAW.

F1 F2 F3 F4 F5 F6

V-WindowEXIT back to

Graph window.DRAW the graph.

Change the window: V-Window; V-MEM; RECALL; 1.

Trace the graph to find y when x = 0.25.

F1 F2 F3 F4 F5 F6

Trace

The value of the function is 1.414213562...

This agrees with the conjecture.

Thus far all fractional exponents have had a numerator of one. Having seen that the laws of exponents hold with those fractional exponents enables a conjecture about fractional exponents whose numerators are not one. This conjecture can be verified using the graphs already created, and then a general definition of a fractional exponent can be created.

Trace the function to find f(3/2), or f(1.5).

This the function used last (Y1 = 4x).

F(1.5) = 8.


F1 F2 F3 F4 F5 F6

V-Window

Change the window: V-Window; V-MEM; RECALL; 2.

F1 F2 F3 F4 F5 F6


Un-SELECT Y1; SELECT Y2.

DRAW.

Trace to find f(2/3), or f(0.6666666667).

F(2/3), or f(0.6666666667), = 4


F1 F2 F3 F4 F5 F6

EXIT to the Graph window; un-select Y2.

SELECT Y4; DRAW.

F1 F2 F3 F4 F5 F6

SELECT Y5; DRAW.

EXIT to the Graph window; un-select Y4.

F1 F2 F3 F4 F5 F6

This graph does exist.

EXIT back to the graph window.

F1 F2 F3 F4 F5 F6

Keep Y5; re-SELECT Y1; DRAW.

The graphs appear to be reflections across x = 0.

If so and the point (a, b) were on one graph, the point (-a, b) would be on the other.

Part II:Exploring

Logarithms

What are logarithms, and how do they relate to the previous exploration?

The logarithmic expression “logb n = x” is read “the logarithm, in base b, of the number n is x.”

As was true for exponential functions, the base of a logarithm can be any positive number.

When a logarithm has a base of 10, the base is not written. Thus, “log x” means the same as “log10 x” and known as the common logarithm.

F1 F2 F3 F4 F5 F6

EXIT to the Graph window; unselect Y1 and Y5.

Select Y6, Y7, and Y8.

F1 F2 F3 F4 F5 F6

Change the view window. With no graph,

SHIFT ; F3.

Change to INITIAL. X,θ,T

F1 F2 F3 F4 F5 F6

X,θ,T

F1 F2 F3 F4 F5 F6


X,θ,T

F1 F2 F3 F4 F5 F6

DRAW the Graphs.

X,θ,TThere appears to be a reflection across y = x.

X,θ,TIf so and a point (a, b) were found on y = 10x, the point (b, a) would be found on y = log x.

X,θ,T

X,θ,T

Discuss what is happening on the graph of y = 10x as the graph takes on x-values that are negative numbers having increasing absolute values.

X,θ,T

X,θ,T

If the graphs of y = 10x and y = log x were reflections in the line y = x, will the graph of y = log x ever touch the y-axis?

Three points that are known to be on the graph of y = 10x are (-1, 0.1), (0, 1), and (1, 10). Therefore, the points (0.1, -1), (1, 0), and (10, 1) should all be on the graph of y = log x.

TRACE. Cursor down twice to access the graph of y = log x. The function is not defined for x = 0, so an error message appears. Cursor right to verify that the three points above are found.

Previously Known Now Established

10-1 = 0.1 log 0.1 = -1

100 = 1 log 1 = 0

101 = 10 log 10 = 1

The logarithm (in base 10) of 0.1 is -1.The logarithm (in base 10) of 1 is 0.The logarithm (in base 10) of 10 is 1.

Previously Known Now Established

10 -1 = 0.1 log 0.1 = -1

10 0 = 1 log 1 = 0

10 1 = 10 log 10 = 1

These are the logarithms (in base 10).What are logarithms, andwhat is the connection between exponential and logarithmic expressions?

4-2 = 0.0625 log4 0.0625

4-1 = 0.25 log4 0.25

40 = 1 log4 1

41 = 4 log4 4

42 = 16 log4 16

Previously Known Predict These

Verify the predictions: MENU; 1

F1 F2 F3 F4 F5 F6

Select the MATH operations .

F1 F2 F3 F4 F5 F6

Select logab (logab).

F1 F2 F3 F4 F5 F6

Enter a base of 4.

Cursor right and enter 0.0625; EXE.

F1 F2 F3 F4 F5 F6

Find log4(0.25).

F1 F2 F3 F4 F5 F6

Find log4(1).

F1 F2 F3 F4 F5 F6

Find log4(4).

F1 F2 F3 F4 F5 F6

Find log4(16).

F1 F2 F3 F4 F5 F6

4-2 = 0.0625 log4 0.0625

4-1 = 0.25 log4 0.25

40 = 1 log4 1

41 = 4 log4 4

42 = 16 log4 16


4-2 = 0.0625 log4 0.0625 = -2

4-1 = 0.25 log4 0.25

40 = 1 log4 1

41 = 4 log4 4

42 = 16 log4 16


4-2 = 0.0625 log4 0.0625 = -2

4-1 = 0.25 log4 0.25 = -1

40 = 1 log4 1

41 = 4 log4 4

42 = 16 log4 16


4-2 = 0.0625 log4 0.0625 = -2

4-1 = 0.25 log4 0.25 = -1

40 = 1 log4 1 = 0

41 = 4 log4 4

42 = 16 log4 16


4-2 = 0.0625 log4 0.0625 = -2

4-1 = 0.25 log4 0.25 = -1

40 = 1 log4 1 = 0

41 = 4 log4 4 = 1

42 = 16 log4 16


4-2 = 0.0625 log4 0.0625 = -2

4-1 = 0.25 log4 0.25 = -1

40 = 1 log4 1 = 0

41 = 4 log4 4 = 1

42 = 16 log4 16 = 2


Logarithms are exponents!

Expression Predict x

log2 32 = x

logx 9 = 1

log5 x = 3

log 100 = x

logx 2 = 2

log⅓ x = -3


log2 32 = x 2x = 32; x = 5

logx 9 = 1

log5 x = 3

log 100 = x

logx 2 = 2

log⅓ x = -3


log2 32 = x 2x = 32; x = 5

logx 9 = 1 x1 = 9; x = 9

log5 x = 3

log 100 = x

logx 2 = 2

log⅓ x = -3


log2 32 = x 2x = 32; x = 5

logx 9 = 1 x1 = 9; x = 9

log5 x = 3 53 = x; x = 125

log 100 = x

logx 2 = 2

log⅓ x = -3


log2 32 = x 2x = 32; x = 5

logx 9 = 1 x1 = 9; x = 9

log5 x = 3 53 = x; x = 125

log 100 = x 10x = 100; x = 2

logx 2 = 2

log⅓ x = -3

Verify predictions using the calculator.

Does log232 = 5?

Does log99 = 1?

Does log5125 = 3?

Does log10100 = 2?

Does log√22 = 2?

Does log⅓27 = -3?

Laws of Logarithms

The calculator can be used to explore patterns involved in logarithmic arithmetic. These patterns will lead to the development of the three laws of logarithms. If the connection between parallel exponential and logarithmic expressions has been grasped, the laws of logarithms will not be surprising.

First Law of Logarithms

log2 (4 × 8)


log2 (4 × 8)

5


log2 (4 × 8) log2 (4)

5


log2 (4 × 8) log2 (4)

5 2


log2 (4 × 8) log2 (4) log2 (8)

5 2


log2 (4 × 8) log2 (4) log2 (8)

5 2 3


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25)


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25)

3


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5)

3


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5)

3 1


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25)


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25)

-1.5


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5)

-1.5


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5)

-1.5 -0.5


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5) log4 (0.25)

-1.5 -0.5


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5) log4 (0.25)

-1.5 -0.5 -1


log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5) log4 (0.25)

-1.5 -0.5 -1


First Law of Logarithms: logb (m×n) = logb m ? logb

n

log2 (4 × 8) log2 (4) log2 (8)

5 2 3

log5 (5 × 25) log5 (5) log5 (25)

3 1 2

log4 (0.5 × 0.25) log4 (0.5) log4 (0.25)

-1.5 -0.5 -1


First Law of Logarithms: logb (m×n) = logb m + logb

n

Second Law of Logarithms

log10 (1000 ÷ 10)


log10 (1000 ÷ 10)

2


log10 (1000 ÷ 10) log10 1000

2


log10 (1000 ÷ 10) log10 1000

2 3


log10 (1000 ÷ 10) log10 1000 log10 10

2 3


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243)


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243)

-3


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243) log3 9

-3


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243) log3 9

-3 2


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243) log3 9 log3 243

-3 2


log10 (1000 ÷ 10) log10 1000 log10 10

2 3 1

log3 (9 ÷ 243) log3 9 log3 243

-3 2 5









Second Law of Logarithms: logb (m ÷ n) = logb m ? logb

n


Second Law of Logarithms: logb (m ÷ n) = logb m – logb

n

Third Law of Logarithms

log9 (32)


log9 (32)

1


log9 (32) power

1


log9 (32) power

1 2


log9 (32) power log9 3

1 2


log9 (32) power log9 3

1 2 1/2
















Laws of Exponents Laws of Logarithms

Enjoy! Kenn

Documents

Technology: A Portal to Exploration and Discovery GCTM October 18th, 2012