How do you solve $log x^2 = (log x)^2$ ?

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Answer 1 · 2015-12-01T21:59:46

$x=100$

Rewrite as:

$2logx=(logx)^2$

$0=(logx)^2-2logx$

$0=logx(logx-2)$

Set each portion equal to $0$ .

$=>logx=0$

$x$ does not exist.

$=>logx-2=0$

$logx=2$

$log_10x=2$

$10^(log_10x)=10^2$

$x=100$

How do you solve log x^2 = (log x)^2log x^2 = (log x)^2?