How do you solve ${log}_{8} 25 = 2 {log}_{8} x$ $log_8 25= 2log_8 x$ ?

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Answer 1 · 2016-04-11T17:59:48

$x = 5$ $x = 5$

Rearrange using laws of logarithms, where a coefficient outside a $log$ $log$ is the same as a power inside it.

${log}_{8} 25 = 2 {log}_{8} x = {log}_{8} x^{2}$ $log_8 25 = 2log_8 x = log_8 x^2$

Raise both sides by eight,

$8^{{log}_{8} 25} = 8^{{log}_{8} x^{2}}$ $8^(log_8 25) = 8^(log_8 x^2)$
$25 = x^{2}$ $25 = x^2$

which gives $x = - 5, 5$ $x = -5, 5$ .

But you can't have a $log$ $log$ of a negative number, which just leaves $5$ $5$ .

How do you solve log825=2log8xlog_8 25= 2log_8 x?