# How do you solve ln(x + 5) + ln(x - 1) = 2?

Sep 21, 2015

$x = - 2 + \sqrt{{e}^{2} + 9}$

#### Explanation:

ln(x+5)+ln(x−1)=2
First notice the domain is:$x > 1$, next:
Simplify the left side(LS) using: $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$ :
$\ln \left[\left(x + 5\right) \left(x - 1\right)\right] = 2$
Expand and simplify inside the bracket on LS:
$\ln \left({x}^{2} + 4 x - 5\right) = 2$
Use the log property: ${\log}_{a} \left(x\right) = y \Leftrightarrow x = {a}^{y}$:
${x}^{2} + 4 x - 5 = {e}^{2}$
Solve the quadratic by completing the square:
${x}^{2} + 4 x = {e}^{2} + 5$
${x}^{2} + 4 x + {\left(\frac{4}{2}\right)}^{2} = {e}^{2} + 5 + {\left(\frac{4}{2}\right)}^{2}$
Simplify:
${x}^{2} + 4 x + 4 = {e}^{2} + 5 + 4$
Or:
${\left(x + 2\right)}^{2} = {e}^{2} + 9$
Take square root of both sides:
$x + 2 = \pm \sqrt{{e}^{2} + 9}$
Subtract 2 from both sides:
$x = - 2 \pm \sqrt{{e}^{2} + 9}$
Reject the negative root (not in the domain):
$x = - 2 + \sqrt{{e}^{2} + 9}$