# How do you solve log_2(x^2 - 10)= log_2 (3x)?

Apr 8, 2016

$x = 5$

#### Explanation:

${\log}_{b} p = {\log}_{b} q$
$\Rightarrow p = q$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {\log}_{2} \left({x}^{2} - 10\right) = {\log}_{2} \left(3 x\right)$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow {x}^{2} - 10 = 3 x$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} - 3 x - 10 = 0$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \left(x - 5\right) \left(x + 2\right) = 0$

$\textcolor{w h i t e}{\text{XXX")rarr x=5color(white)("XX")orcolor(white)("XX}} x = - 2$

But neither $\log$ is defined when $x = - 2$ and therefore this result is extraneous.