We have the function f(x) = e^(3x - 5).

Let y = e^(3x - 5).

=> y = e^3x / e^5

Multiply both sides by e^5

=> y*e^5 = e^3x

take the log to the base e of both the sides

=> ln ( y*e^5) = 3x

=> x =...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We have the function f(x) = e^(3x - 5).

Let y = e^(3x - 5).

=> y = e^3x / e^5

Multiply both sides by e^5

=> y*e^5 = e^3x

take the log to the base e of both the sides

=> ln ( y*e^5) = 3x

=> x = (ln y + ln e^5)/3

=> x = (ln y + 5)/3

Interchange y and x

=> y = (ln x + 5)/3

**Therefore the inverse of f(x) = e^(3x - 5) is f(x) = (ln x + 5)/3**

Given the function f(x) = e^(3x - 5). We need to find the inverse (f^-1 (x) of the function f(x).

Let y= e^(3x-5).

Now we will apply the natural logarithm for both sides.

==> ln y = ln e^(3x-5).

Now we will use the logarithm properties to simplify the equation.

We know that ln e^a = a*lne = a*1 = a

==> ln y = (3x-5)*ln e

==> ln y = (3x -5) *1

==> ln y = (3x-5)

Now we will add 5 to both sides.

==> 5 + ln y = 3x

Now we will divide by 3.

==> (5+ln y)/3 = x

==> x = (5+ln y)/ 3

Now we will rewrite x as y and y as x.

==> y= (5+ln x)/3

Then the inverse function is:

**f^-1 (x) = (5+ ln x)/3**