# 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.

$$x^{\sqrt[\leftroot{-2}\uproot{2} x]{2}} = 2$$

• You have: $$x^{2^{1/x}}=2$$ or, flipping it all upside down a bit, $$\frac12=y^{2^y}$$ where $y=\frac1x$. Clearly, this is not solvable... at least not in any algebraic sort of manner. – Simply Beautiful Art Apr 14 '17 at 20:40

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 $(1,2)$.
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 $$\left( \begin{array}{cc} n & t_n \\ 0 & 0.5000000000 \\ 1 & 0.6336043641 \\ 2 & 0.6411741260 \\ 3 & 0.6411857445 \end{array} \right)$$