# How do you solve log_2 (x+2) - log_2 (x-5) = 3?

Jan 2, 2016

Unify the logarithms and cancel them out with ${\log}_{2} {2}^{3}$

$x = 6$

#### Explanation:

${\log}_{2} \left(x + 2\right) + {\log}_{2} \left(x - 5\right) = 3$

Property $\log a - \log b = \log \left(\frac{a}{b}\right)$

${\log}_{2} \left(\frac{x + 2}{x - 5}\right) = 3$

Property $a = {\log}_{b} {a}^{b}$

${\log}_{2} \left(\frac{x + 2}{x - 5}\right) = {\log}_{2} {2}^{3}$

Since ${\log}_{x}$ is a 1-1 function for $x > 0$ and $x \ne 1$, the logarithms can be ruled out:

$\frac{x + 2}{x - 5} = {2}^{3}$

$\frac{x + 2}{x - 5} = 8$

$x + 2 = 8 \left(x - 5\right)$

$x + 2 = 8 x - 8 \cdot 5$

$7 x = 42$

$x = \frac{42}{7}$

$x = 6$