# How do I solve for x in ln &#x2061;<!-- ⁡ --> ( x ) ln &#x2061;<!-- ⁡ -->

How do I solve for x in $\mathrm{ln}\left(x\right)\mathrm{ln}\left(x\right)=2+\mathrm{ln}\left(x\right)$
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HINT:
Putting $\mathrm{ln}\left(x\right)=y,$ , we get
${y}^{2}-y-2=0.$
Do you know how to solve a Quadratic Equation?, and we know that
$\mathrm{ln}\left(x\right)=a\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x={e}^{a}.$
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pouzdrotf
$\begin{array}{}\text{(1)}& \mathrm{ln}\left(x\right)\mathrm{ln}\left(x\right)=2+\mathrm{ln}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{ln}\left(x\right){\right)}^{2}-\mathrm{ln}\left(x\right)-2=0\end{array}$
Let $t=\mathrm{ln}x$ . Then (1) becomes
$\begin{array}{}\text{(2)}& {t}^{2}-t-2=\left(t-2\right)\left(t+1\right)=0\end{array}$
So,