# How do you solve  log_6 (x+ 1)+ log_6 (x-4)= 1 ?

Jul 30, 2016

$x = 5$

#### Explanation:

Given:

${\log}_{6} \left(x + 1\right) + {\log}_{6} \left(x - 4\right) = 1$

We require:

$6 = {6}^{1} = {6}^{{\log}_{6} \left(x + 1\right) + {\log}_{6} \left(x - 4\right)} = \left(x + 1\right) \left(x - 4\right) = {x}^{2} - 3 x - 4$

Subtract $6$ from both ends to get:

$0 = {x}^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right)$

So $x = 5$ or $x = - 2$

If $x = 5$ then:

${\log}_{6} \left(x + 1\right) + {\log}_{6} \left(x - 4\right) = {\log}_{6} \left(6\right) + {\log}_{6} \left(1\right) = 1 + 0 = 1$

So $x = 5$ is a solution of the original equation.

If $x = - 2$ then (even if we allow Complex logarithms):

${\log}_{6} \left(x + 1\right) + {\log}_{6} \left(x - 4\right)$

$= {\log}_{6} \left(- 1\right) + {\log}_{6} \left(- 6\right)$

$= {\log}_{6} \left(1\right) + \frac{\pi i}{\ln} \left(6\right) + {\log}_{6} \left(6\right) + \frac{\pi i}{\ln} \left(6\right)$

$= 1 + \frac{2 \pi i}{\ln} \left(6\right) \ne 1$

So $x = - 2$ is not a solution of the original equation.