How do you solve ${log x}^{2} = 2$ $log x^2 =2$ ?

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Answer 1 · 2015-10-05T00:10:20

$x = \pm 10$ $x = +-10$

Assuming that's the common log, take the base 10 exponential of both sides, that is, the "sideth" power of 10.

$log (x^{2}) = 2$ $log(x^2) = 2$
$x^{2} = 10^{2}$ $x^2 = 10^2$

Take the root

$x = \pm 10$ $x = +-10$

Since $x^{2}$ $x^2$ is positive for all values of $x$ $x$ the only value of $x$ $x$ we can't have is $0$ $0$ , so both answers are okay.

How do you solve log x^2 =2logx2=2log x^2 =2?