Exponential Functions to Logarithmic Functions: Math 30 Pure

Consider the function below: y

2(3) x

4

Domain: x R Range: y > 4, y

R

Determine the equation of the inverse of y

2(3) x

4.

Step 1: Swap “ x ” and “ y ” x

2(3) y

4

Step 2: Isolate “ y ” x 4 2(3) y 1 ( x 4) 3 y 2

Range: y R Domain: x > 4, x R

Solve for the x-intercept algebraically. Set y

0 in the equation y

0 log3 1 (x 2 1 (x 2 x 4 x 4 x 6

1 ( x 4) 2

log 3

1 ( x 4) 2

and solve for “ x ”.

 convert to exponential form…

4) 30 4) 1

2 2 

The x-intercept in this example is 6, which corresponds to the y-intercept in the original function y 2(3) x 4 .

Consider another example:

Solve for y-intercept: y

y

y

3 0 (2) 6 2 3 6 2 4.5

Solve for x-intercept: 3 x (2) 6 2 3 x 6 (2) 2 4 2x x 2 0

Determine the equation of the inverse: y

3 x (2) 6  x 2

3 y (2) 6  x 6 2

Solve for x-intercept: 0 log 2 20

1 3 2

2 ( x 6) 3

2 ( x 6) 3

2 ( x 6) 3 x 6

x

6

x

4.5

3  2

3 y 2 (2)  ( x 6) 2 3

2y  y

Solve for y-intercept: 2 (6) 3

y

log 2

y

log 2 4

y

2

log 2

2 ( x 6) 3

Conclusions:

1 ( x 10) 7 vertical asymptote: x 10 domain: x > 10, x R

Given the function y

log 2

The x-intercept should be 7 units to the right of the vertical asymptote. This is referred to as a horizontal stretch about the line x 10 when being compared to the graph defined by the function y log 2 ( x 10) . Proof: Solve for the x-intercepts for the following:

log 2

1 ( x 10) 7

0 log 2 ( x 10)

0 log 2

1 ( x 10) 7

20

20

y

y

log 2 ( x 10)

x 10

1 ( x 10) 7 7 x 10

1

1 x 10

x

1 ( x 10) 7

9 x

3

One more example to consider:

y

log 2

1 ( x 3) 5

vertical asymptote: x 3

From what we’ve discovered so far, the x-intercept should be 5 units to the left of x 3 .

Verify:

0 log 2

1 ( x 3) 5

20 1

x

The negative sign on

1 ( x 3) 5

1 ( x 3) 5 5 x 3

2

1 caused a reflection in the line x 3 . 5