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

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Answer 1 · 2018-01-16T17:09:16

$x = \pm 4$ $x=+-4$

${log}_{8} 2 + {log}_{8} 2 x^{2} = {log}_{8} 64$ $log_8 2+log_8 2x^2=log_8 64$

Rearrange:

$\to {log}_{8} 2 x^{2} = {log}_{8} 64 - {log}_{8} 2$ $->log_8 2x^2=log_8 64 - log_8 2$

Using the rules of logs:

${log}_{8} 2 x^{2} = {log}_{8} (\frac{64}{2})$ $log_8 2x^2=log_8 (64/2)$

${log}_{8} 2 x^{2} = {log}_{8} (32)$ $log_8 2x^2=log_8 (32)$

So:

$8^{{log}_{8} 2 x^{2}} = 8^{{log}_{8} (32)}$ $8^(log_8 2x^2)=8^(log_8 (32))$

$cancel8^(cancel(log_8) 2x^2)=cancel8^(cancel(log_8) (32))$

$->2x^2=32$

$x^2=16-> x=+-4$

How do you solve \log _ { 8} 2+ \log _ { 8} 2x ^ { 2} = \log _ { 8} 64log82+log82x2=log864\log _ { 8} 2+ \log _ { 8} 2x ^ { 2} = \log _ { 8} 64?