Answer
$x=2.718$
Work Step by Step
$\frac{1-\ln x}{x^2}=0$
$1-\ln x=0$
$1=\ln x$
$e^{1}=e^{\ln x}$
$x=2.718$
Notice that $2.718 $ lies in the domain of $\ln x$ and in the domain of $\frac{1-\ln x}{x^2}$
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