How do you solve for x in ${ln x}^{2} = 4$ $Lnx^2=4$ ?

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Answer 1 · 2016-07-11T11:02:38

$x \pm e^{2}$ $x ± e^2$

Taking the natural logarithm of both sides gives us

$cancel(e)^(cancel(ln)(x^2)) = e^(4)$

$x^2 = e^4$

Taking the square root of both sides yields

$x ± sqrt(e^4)$

Simplifying further then gives us our result of

$x ± e^2$

How do you solve for x in Lnx^2=4lnx2=4Lnx^2=4?