Find the inverse of the function f(x) = e^(3x - 5).

### 5 Answers | Add Yours

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**

We'll put the function y=f(x)

y = e^(3x - 5)

We'll take natural logarithms both sides:

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

We'll use the power rule of logarithms:

ln y = (3x - 5)ln e

But ln e = 1

ln y = 3x - 5

We'll use the symmetric property:

3x - 5 = ln y

We'll isolate x to the left side:

3x = ln y + 5

x= (ln y + 5)/3

**The inverse function is:**

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

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**

To find the inverse of f(x) = e^(3x-5).

Let the inverse of f(x) be y = f^-1 (x) .

Then by definition of inverse functions, f(y) = x.

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

Therefore we have to solve for y to get the inverse:

Taking logarithms to the base e, we get:

lnx = 3y-5

Or 5+lnx = 3y.

Or (5+lnx)/3 = y.

Threfore y = (1/3) (5+lnx) is the inverse of f(x) = e^(3x-5)

Any function f(x) and its inverse `f^-1(x)` follow the relation: `f(f^-1(x)) = x` .

For the function `f(x) = e^(3x - 5)` , to determine the inverse, we use the relation `f(f^-1(x)) = x` .

`f(f^-1(x)) = x`

`e^(3*f^-1(x) - 5) = x`

Now take the logarithm to base e for both the sides.

`log_e (e^(3*f^-1(x) - 5)) = log_e x`

Use the relation `log a^b = b*log a` and `log_b b = 1`

`(3*f^-1(x) - 5)*log_e e = log_e x`

`3*f^-1(x) - 5 = log_e x`

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

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

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

The required inverse function is `f^-1(x) = (5 + ln x)/3`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes