How do you solve for x in ${log 10}^{x^{2}} = 4$ $log10^(x^2)=4$ ?

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Answer 1 · 2016-01-13T07:43:18

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

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

So:

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

Hence $x = \pm 2$ $x = +-2$

How do you solve for x in log10x2=4log10^(x^2)=4?