# Find `y=f(f^(-1)(3x))` in the form of y=ax/(bx + c), where a,b and c are real constants. `f(x)=2log_e(x+1)` and ` f^(-1)(x)=e^(x/2) - 1` .

### 2 Answers | Add Yours

We have

`f(x) = 2log_e(x+1)` and

`f^(-1)(x) = e^(x/2)-1`

To work out

`y = f(f^(-1)(3x))`

however, we don't need either of these as, since `f^(-1)(x)`is the inverse of `f(x)`,

`f(f^(-1)(3x))` is simply `3x`.

We can write `y = f(f^(-1)(3x))` in the form `(ax)/(bx+c)` where

`a = 3`, `b = 0` and `c=1`

**y = 3x/(0x + 1)**

`f(x)=2log_e(x+1)`

`f^(-1)(x)=e^(x/2)-1`

`f(f^(-1)(3x))=f(e^((3x)/2)-1)`

`=2log_e(e^((3x)/2)-1+1)`

`f(f^(-1)(3x))=2xx((3x))/2`

`f(f^(-1)(3x))=3x`

`y=f(f^(-1)(3x))=3x`

a=1,b=0,c=1

`y=(1.x)/(0.x+1)`