# How do you find the inverse of y = ln(x) + ln(x-6)?

Dec 4, 2015

Solve for $x$ using properties of logarithms and the quadratic formula and eliminate an extraneous solution to find
${f}^{-} 1 \left(x\right) = 3 + \frac{\sqrt{36 - 4 {e}^{x}}}{2}$

#### Explanation:

We'll proceed under the assumption you are trying to find the inverse of the function $f \left(x\right) = \ln \left(x\right) + \ln \left(x - 6\right)$

In general, to find the inverse of a function, a good method is to set $y = f \left(x\right)$ and then solve for $x$ to obtain $x = {f}^{- 1} \left(y\right)$
(To see why this works, substitute in $f \left(x\right)$ for $y$ and note that the result is ${f}^{- 1} \left(f \left(x\right)\right) = x$ as desired.)

To do that here, we will need to use the following:

• $\ln \left(a\right) + \ln \left(b\right) = \ln \left(a b\right)$
• ${e}^{\ln} \left(a\right) = a$
$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Let $y = f \left(x\right) = \ln \left(x\right) + \ln \left(x - 6\right)$

(note here that as we have $\ln \left(x - 6\right)$ it must be that $x > 6$)

$\implies y = \ln \left(x \left(x - 6\right)\right) = \ln \left({x}^{2} - 6 x\right)$

$\implies {e}^{y} = {e}^{\ln \left({x}^{2} - 6 x\right)} = {x}^{2} - 6 x$

$\implies {x}^{2} - 6 x - {e}^{y} = 0$

$\implies x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(1\right) \left(- {e}^{y}\right)}}{2 \left(1\right)}$

$= \frac{6 \pm \sqrt{36 + 4 {e}^{y}}}{2}$

$= 3 \pm \frac{\sqrt{36 + 4 {e}^{y}}}{2}$

But $3 - \frac{\sqrt{36 + 4 {e}^{y}}}{2} < 6$, so, as noted above, we must throw it out as a possible solution for $x$

Thus

$x = 3 + \frac{\sqrt{36 - 4 {e}^{y}}}{2}$

Then, by our process, we have

${f}^{-} 1 \left(x\right) = 3 + \frac{\sqrt{36 - 4 {e}^{x}}}{2}$