Solving an equation involving the x to the xth root of 2 How do I solve the following equation for $ x $? Using wolfram alpha gives me an answer, but not the steps, and the step-by-step solvers I tried couldn't solve this.
$$ x^{\sqrt[\leftroot{-2}\uproot{2} x]{2}} = 2 $$
 A: The equation can be written
$$
\sqrt[x]{2}\log x=\log 2
$$
(natural logarithm); setting $t=1/x$, it becomes
$$
-2^t\log t=\log 2
$$
Consider the function $f(t)=2^t\log t+\log 2$. We see that
$$
\lim_{t\to0}f(t)=-\infty
\qquad
\lim_{t\to\infty}f(t)=\infty
$$
Moreover
$$
f'(t)=2^t\cdot\log 2\cdot\log t+\frac{2^t}{t}>0
$$
so the function is increasing. Then the equation $f(t)=0$ has exactly one solution.
Since $f(1)=\log2>0$ and
$$
f(1/2)=-\sqrt{2}\log2+\log2=(1-\sqrt{2})\log2<0,
$$
we know the solution is in the interval $(1/2,1)$, so the solution of the original equation is in $(1,2)$.
A: Alpha only gives a numeric approximation which, unless you mis-typed your equation into alpha, should be about 1.56.
There is no simple closed for for the answer.
But there is something cute about this:  Full Mathematica fails to give a numerical solution using its NSolve method, while Alpha does give a numerical solution -- even when I type it in using "NSolve[x^(2^(1/x))==2,x]" (!)
A: Starting using egreg's solution, consider that you look for the zero of  function $$f(t)=2^t\log t+\log 2$$ $$f'(t)=\frac{2^t}{t}+2^t \log (2) \log (t)$$ and, starting from a "reasonable" guess $t_0$, Newton method will update it according to $$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}$$ Starting with $t_0=1$, the successive iterates would then be
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 1.000000000 \\
 1 & 0.6534264097 \\
 2 & 0.6411588810 \\
 3 & 0.6411857444 \\
 4 & 0.6411857445
\end{array}
\right)$$
Starting with $t_0=\frac 12$, the successive iterates would then be
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 0.5000000000 \\
 1 & 0.6336043641 \\
 2 & 0.6411741260 \\
 3 & 0.6411857445
\end{array}
\right)$$
