# What is the inverse function for f(x) = 3*ln (2x-1) + 5.

Asked on by ryanomar

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function f(x) = 3*ln (2x - 1) + 5

let y = f(x)

=> y = 3*ln (2x - 1) + 5

=> y - 5 = 3*ln (2x - 1)

=> (y - 5)/3 = ln (2x - 1)

=> 2x - 1 = e^[(y - 5)/3]

=> 2x = e^[(y - 5)/3] + 1

=> x = (e^[(y - 5)/3] + 1)/2

interchange x and y

=> y = (e^[(x - 5)/3] + 1)/2

The inverse of f(x) is f^-1(x) = (e^[(x - 5)/3] + 1)/2

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(x) = 3*ln (2x-1) + 5

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

let y= 3*ln (2x-1) + 5

We need to isolate x by itself.

First we will subtract 5 from both sides.

==> y-5 = 3ln (2x-1)

Now we will divide by 3.

==> (y-5)/3 = ln (2x-1)

Now we will rewrite into the exponent form.

==> 2x-1 = e^[(y-5)/3]

Now we will add (1).

==> 2x = e^[(y-5)/3] + 1

Now we will divide by 2.

==> x = { e^[(y-5)/3] + 1 }/ 2

==> f^-1 (x) = { e^[(x-5)/3] + 1 }/ 2

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