Question

Find the inverse of the function? : `h(x) = log((x+9)/(x−6))`

Answer 1

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

Explanation:

We have:

`h(x) = log((x+9)/(x−6))`

To find `h^(-1)(x)` we need to rearrange to the equation to the for `x=f(h)`.

Writing as:

`h = log((x+9)/(x−6))`

`:. (x+9)/(x−6) = e^h`

`:. x+9 = (x−6)e^h`

`:. x+9 = xe^h−6e^h`

`:. xe^h - x = 6e^h + 9`

`:. x(e^h - 1) = 3(2e^h + 3)`

`:. x = 3 ( (2e^h + 3) / (e^h - 1) )`

Hence, the inverse function is:

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

I have assumed natural logarithms (base e). If base `10` logarithms, just change `e` to `10`

Answer 2

`y = (9+ (6*10^x))/(10^x - 1)`

Explanation:

To find the inverse, let us switch the x and y variables, denoting `h(x)` as `y`

`y = log((x+9)/(x-6))` `implies` `x = log((y+9)/(y-6))`

Assuming `log(x) = log_10 (x),`

Adding a base 10 to each side of the equation to cancel out the `log`, since `10^log(x) = x`

`10^x = 10^(log((y+9)/(y-6)))`

`10^x = (y+9)/(y-6)`

Multiplying both sides by `y-6` and subtracting `y`

`10^x(y-6) = y+9` `implies` `10^xy - (6*10^x)-y = 9`

Taking all `y` terms to one side:

`10^xy-y = 9 + (6*10^x)`

`y(10^x - 1) = 9+ (6*10^x)`

`y = (9+ (6*10^x))/(10^x - 1)`

I am aware of the other variations in which this answer could be rewritten, but you can work off of this answer to your preference.
Hope this helped!