# Find the inverse of the function f(x) = -2ln(2x-3) + 2. We have the function f(x) = -2ln(2x-3) + 2

Let y = f(x) = -2*ln (2x - 3) + 2

=> y - 2 = -2*ln (2x - 3)

=> -(y-2)/2 = ln (2x - 3)

=> e^ ( (2-y)/2)) = 2x - 3

=> 2x = 3 + e^(...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

We have the function f(x) = -2ln(2x-3) + 2

Let y = f(x) = -2*ln (2x - 3) + 2

=> y - 2 = -2*ln (2x - 3)

=> -(y-2)/2 = ln (2x - 3)

=> e^ ( (2-y)/2)) = 2x - 3

=> 2x = 3 + e^( (2-y)/2))

=> x = [3 + e^( (2-y)/2))]/2

=> x = 3/2 + e^(1 - y/2)/2

interchange x and y , we get the inverse function as f(x) = 3/2 + e^(1 - x/2)/2.

Therefore the required inverse is f(x) = 3/2 + e^(1 - x/2)/2.

Approved by eNotes Editorial Team Given the function f(x) = -2ln(2x-3) + 2

We need to find the inverse function f^-1 (x).

Let us assume that f(x) = y

==> y= -2ln(2x-3) + 2

The goal is to isolate x on one side .

Let us subtract 2 from both sides.

==> y-2 = -2ln(2x-3)

Now we will divide by -2

==> (y-2)/-2 = ln (2x-3)

Now we will rewrite using the exponent form:

==> 2x-3 = e^(2-y)/2

Now we will add 3 to both sides.

==> 2x = e^(2-y)/2  + 3

Now we will divide by 2.

==> x = [ e^(2-y)/2  + 3] / 2

Now we will replace y and x.

==> y= [ e^(2-y)/2  + 3 ] / 2

Then the inverse function is f^-1 (x) = [ e^(2-y)/2  + 3] / 2

Approved by eNotes Editorial Team