# 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` .

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### 2 Answers

You need to find first `f^(-1)(3x)` replacing `3x` for `x ` in equation of `f^(-1)(x)` , such that:

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

You need to find now `f(f^(-1)(3x))` replacing `f^(-1)(3x)` for x in equation of `f(x)` , such that:

`f(f^(-1)(3x)) = 2 ln (f^(-1)(3x) + 1)`

Replacing `e^((3x)/2) - 1` for `f^(-1)(3x)` yields:

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

Reducing like terms yields:

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

Using the following logarithmic identity `ln a^b = b*ln ` a, yields:

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

Since `ln e = 1` `=> f(f^(-1)(3x)) = 3x`

Since you need to put `f(f^(-1)(3x)) = y` in the form `y = (ax)/(bx+c)` , comparing the expression yields:

`3x = (ax)/(bx+c) => {(a = 3),(b = 0),(c = 1):}`

**Hence, evaluating the function y, under the given conditions, yields:**`y = (3x)/(0*x + 1)`

Since we know `f(f^(-1)(x))=x` ( by definition of inverse)

Therefore

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

`y=(ax)/(bx+c)` (ii)

from (i) and (ii)

a=3 , bx+c=1

a=3, c=1 and b=0

only possible solution.

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