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

Jun 21, 2016

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

e = 2.7183, nearly.

#### Explanation:

$x > 1$ to make $\log \left(x - 1\right)$ real.

Use $\ln {e}^{n} = n \mathmr{and} \ln a + \ln b = \ln a b$..

The given equation is

$\ln \left(\left(x + 5\right) \left(x - 1\right)\right) = \ln {e}^{2}$.

So, $\left(x + 5\right) \left(x - 1\right) = {e}^{2}$. Solving,

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

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