# How do you solve \log _ { 8} 2+ \log _ { 8} 2x ^ { 2} = \log _ { 8} 64?

Jan 16, 2018

$x = \pm 4$

#### Explanation:

${\log}_{8} 2 + {\log}_{8} 2 {x}^{2} = {\log}_{8} 64$

Rearrange:

$\to {\log}_{8} 2 {x}^{2} = {\log}_{8} 64 - {\log}_{8} 2$

Using the rules of logs:

${\log}_{8} 2 {x}^{2} = {\log}_{8} \left(\frac{64}{2}\right)$

${\log}_{8} 2 {x}^{2} = {\log}_{8} \left(32\right)$

So:

${8}^{{\log}_{8} 2 {x}^{2}} = {8}^{{\log}_{8} \left(32\right)}$

${\cancel{8}}^{\cancel{{\log}_{8}} 2 {x}^{2}} = {\cancel{8}}^{\cancel{{\log}_{8}} \left(32\right)}$

$\to 2 {x}^{2} = 32$

${x}^{2} = 16 \to x = \pm 4$