# How do you solve ln x = 1 - ln (x+8)?

Aug 6, 2015

I found: $x = - 4 + \sqrt{16 + e}$

#### Explanation:

Write it rearranging as:
$\ln \left(x\right) + \ln \left(x + 8\right) = 1$
use the fact that $\ln x + \ln y = \ln \left(x y\right)$
so:
$\ln \left[x \left(x + 8\right)\right] = 1$
use the definition of log:
$\ln x = a \to x = {e}^{a}$
$x \left(x + 8\right) = {e}^{1}$
${x}^{2} + 8 x - e = 0$
${x}_{1 , 2} = \frac{- 8 \pm \sqrt{64 + 4 e}}{2} = \frac{- 8 \pm 2 \sqrt{16 + e}}{2} =$
$= - 4 \pm \sqrt{16 + e}$
so you get two solutions but one, $- 4 - \sqrt{16 + e}$, doesn't work when substituted into the original equation (it is a negative number, try it!) and you can discard it keeping only:
$x = - 4 + \sqrt{16 + e}$

Aug 6, 2015

x=$- 4 \pm \sqrt{16 + e}$

#### Explanation:

Rewriting the equation , it is ln x + ln(x+8) =1

ln x(x+8)=1

${x}^{2} + 8 x = e$

${x}^{2} + 8 x - e = 0$

x= $\frac{- 8 \pm \sqrt{64 + 4 e}}{2}$

x=$- 4 \pm \sqrt{16 + e}$