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

Sep 18, 2017

${h}^{- 1} \left(x\right) = 3 \left(\frac{2 {e}^{x} + 3}{{e}^{x} - 1}\right)$

#### Explanation:

We have:

 h(x) = log((x+9)/(x−6))

To find ${h}^{- 1} \left(x\right)$ we need to rearrange to the equation to the for $x = f \left(h\right)$.

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

$\therefore x {e}^{h} - x = 6 {e}^{h} + 9$

$\therefore x \left({e}^{h} - 1\right) = 3 \left(2 {e}^{h} + 3\right)$

$\therefore x = 3 \left(\frac{2 {e}^{h} + 3}{{e}^{h} - 1}\right)$

Hence, the inverse function is:

${h}^{- 1} \left(x\right) = 3 \left(\frac{2 {e}^{x} + 3}{{e}^{x} - 1}\right)$

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

Sep 18, 2017

$y = \frac{9 + \left(6 \cdot {10}^{x}\right)}{{10}^{x} - 1}$

#### Explanation:

To find the inverse, let us switch the x and y variables, denoting $h \left(x\right)$ as $y$

$y = \log \left(\frac{x + 9}{x - 6}\right)$ $\implies$ $x = \log \left(\frac{y + 9}{y - 6}\right)$

Assuming $\log \left(x\right) = {\log}_{10} \left(x\right) ,$

Adding a base 10 to each side of the equation to cancel out the $\log$, since ${10}^{\log} \left(x\right) = x$

${10}^{x} = {10}^{\log \left(\frac{y + 9}{y - 6}\right)}$

${10}^{x} = \frac{y + 9}{y - 6}$

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

${10}^{x} \left(y - 6\right) = y + 9$ $\implies$ ${10}^{x} y - \left(6 \cdot {10}^{x}\right) - y = 9$

Taking all $y$ terms to one side:

${10}^{x} y - y = 9 + \left(6 \cdot {10}^{x}\right)$

$y \left({10}^{x} - 1\right) = 9 + \left(6 \cdot {10}^{x}\right)$

$y = \frac{9 + \left(6 \cdot {10}^{x}\right)}{{10}^{x} - 1}$

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!