How do you solve for x in $log10^(x^2)=4$ ?

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Answer 1 · 2016-01-13T07:43:18

$log(x)$ and $10^x$ are inverse functions, hence $x^2=4$ so $x=+-2$ .

As Real valued functions, $log(x)$ and $10^x$ are inverses of one another.

So:

$4 = log(10^(x^2)) = x^2$

Hence $x = +-2$

How do you solve for x in log10^(x^2)=4log10^(x^2)=4?