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Is this possible to solve algebraically? I couldn't figure out a way and got approximately 1.5591... and I couldn't see any relationship between that number and any other logarithmic numbers. Is this possible to solve without just guessing and checking?

Arnaldo
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    This equation requires that new functions. Wiki "Lambert's W". The solution to this equation is $x = \frac{\ln 2}{\mbox{LambertW}(\ln 2)}.$ – B. Goddard Aug 16 '17 at 16:36

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Let \begin{align} x^x &= z\\ x\ln x &= \ln z\\ e^{\ln x}\ln x &= \ln z. \end{align} Since for all $t\geq0$ the Lambert W function has the property $$ W\left(t\cdot e^t\right) = t, $$ for $t=\ln x$, we have $W\left(\ln x \cdot e^{\ln x}\right) = \ln x$, so by applying $W$ on both sides, we have $$ \ln x = W\left(\ln z\right). $$ From here we have $$ x = e^{W\left(\ln z\right)}, $$ or since by the definiction of the $W$ function $\ln z = W\left(\ln z\right)e^{W\left(\ln z\right)}$, an equivalent solution is $$ x = \frac{\ln z}{W\left(\ln z\right)}. $$ For $z=2$ this argument answers your question.

user153012
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