One can see that the solutions are $x=2, 4$ and $x=-0.77$(approximately) seen from the graph. I am posting this to find if there is a way to solve this and find solutions like polynomial equations. Thanks !!!
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1Equations that have the unknown in an exponent and in a base can not be solved symbolically. – ajotatxe Feb 05 '16 at 19:46
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see also http://math.stackexchange.com/questions/1544703 – user84413 Feb 05 '16 at 19:49
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The trick here is the Lambert W function, an inverse of $we^w$. We have $$x=\sqrt{2}^x=e^{x\ln\sqrt{2}},\,-x\ln\sqrt{2}e^{-x\ln\sqrt{2}}=-\ln\sqrt{2},\,x=-\frac{-W\left(-\ln\sqrt{2}\right)}{\ln\sqrt{2}}.$$ That's the theory, anyway. Like complex exponentiation, the Lambert W-function has multiple branches in $\mathbb{C}$. But the Lambert W function is of interest because many equations' solutions are expressible in terms of it.
J.G.
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