# How do you solve log_4 7 + 2 log_4 x = log_4 2?

Dec 14, 2015

$x = \sqrt{\frac{2}{7}}$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{4} \left(7\right) + 2 {\log}_{4} \left(x\right) = {\log}_{4} \left(2\right)$

Remember
$\textcolor{w h i t e}{\text{XXX}}$log multiplication rule: ${\log}_{b} \left(p \cdot q\right) = {\log}_{b} \left(p\right) + {\log}_{b} \left(q\right)$
$\textcolor{w h i t e}{\text{XXX}}$log power rule: ${\log}_{b} \left({s}^{t}\right) = t \cdot {\log}_{b} \left(s\right)$

Therefore the given equation can be rewritten as
$\textcolor{w h i t e}{\text{XXX}} {\log}_{4} \left(7 {x}^{2}\right) = {\log}_{4} \left(2\right)$

from which it follows that
$\textcolor{w h i t e}{\text{XXX}} 7 {x}^{2} = 2$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = \frac{2}{7}$

$\textcolor{w h i t e}{\text{XXX}} x = \sqrt{\frac{2}{7}}$
$\textcolor{w h i t e}{\text{XXXXXX}}$we can ignore the negative root as extraneous since $\sqrt{x}$ requires $x \ge 0$ for Real solutions