How does one solve $x = \sqrt[x]2$ for $x$? This can be otherwise stated as
$x = 2^{1/x}$
Raising both sides to the power of $x$:
$x^x = (2^{1/x})^x$
$x^x = 2$
But I don't know where I can go from here.
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1You will need to use the Lambert W function : https://en.wikipedia.org/wiki/Lambert_W_function – Bérénice May 31 '16 at 06:39
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2There exists a solution $x>0$, it is unique, it is such that $$x\ln x=\ln2,$$ and it cannot be expressed in terms of usual functions. Lambert W function will provide the essentially useless formula $$\ln x=W(\ln2).$$ – Did May 31 '16 at 06:39
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Alternately, if this is a homework problem, this may be the time to use your calculator's root-finding methods. – Neil May 31 '16 at 06:39
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Purely curiosity – I had already found a numerical solution, but I hoped for an expression. Lambert W function it is; I'll have a look. – Dodo May 31 '16 at 06:54
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There is no closed form solution for this problem. Even the Lambert W function is not an elemenary function. You could express the solution by using fixed point iteration. – MrYouMath May 31 '16 at 13:12