How do you solve $2 ln x = 1$ $2 ln x = 1$ ?

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Answer 1 · 2016-05-03T00:40:42

$x = \sqrt{e}$ $x=sqrt(e)$

$2 ln (x) = 1$ $2 ln(x) = 1$
$ln (x^{2}) = 1$ $ln(x^2) = 1$
$x^{2} = e^{1}$ $x^2 = e^1$
$x = \pm \sqrt{e}$ $x = +-sqrt(e)$

Since $ln (x)$ $ln(x)$ is not defined for $x < 0$ $x<0$
$x = \sqrt{e}$ $x = sqrt(e)$

How do you solve 2lnx=12 ln x = 1?