How can I solve the given equation?

In addition to the answer from Claude Leibovici the analytical solution is :

$$x^{20}\ln(x)=\frac{\ln(2)}{\sqrt{2}}$$ $$x^{20}\ln(x^{20})=20\frac{\ln(2)}{\sqrt{2}}$$ $$X=\ln(x^{20})\quad\implies\quad Xe^X=20\frac{\ln(2)}{\sqrt{2}}$$ $$X=W\left(20\frac{\ln(2)}{\sqrt{2}}\right)$$ $W$ is the LambertW function. http://mathworld.wolfram.com/LambertW-Function.html $$x^{20}=\exp(X)=\exp\left(W\left(20\frac{\ln(2)}{\sqrt{2}}\right)\right)$$ $$x=\exp\left(\frac{1}{20}W\left(20\frac{\ln(2)}{\sqrt{2}}\right)\right)$$ Numerically :

$20\frac{\ln(2)}{\sqrt{2}}\simeq 9.8025814347$

$X=W(20\frac{\ln(2)}{\sqrt{2}})\simeq 1.7328679514$

$x=\exp{(X/20)}\simeq 1.0905077327$

This is in good agreement with the result of Claude Leibivici, obtained with Newton method, which of course is more usable on practical viewpoint ( if the function LambertW isn't implemented in your math. package).

Consider that you look for the zero of function $$f(x)=x^{x^{20}}-2^{\frac{1}{\sqrt{2}}}$$ If you plot around $x=1$ it does not look bad at all and shows a root close to $1.1$.

So, use Newton method and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.10000 \\ 1 & 1.09211 \\ 2 & 1.09056 \\ 3 & 1.09051 \end{array} \right)$$

As an alternative by bisection method for $f(x)=x^{x^{20}}-2^{\frac{1}{\sqrt{2}}}$ we have

• $a_1=1 \implies f(a_1)=1-2^{\frac{1}{\sqrt{2}}}<0$

• $b_1=1.1 \implies f(b_1)=1-2^{\frac{1}{\sqrt{2}}}>0$

then consider

• $c_1=\frac{a_1+b_1}{2}=1.05 \implies f(c_1)=(1.05)^{(1.05)^{20}}-2^{\frac{1}{\sqrt{2}}}>0$

then let $a_2=a_1$ and $b_2=c_1$ and repeat the process taking at each step $a_{i+1}=a_i$ and $b_{i+1}=c_i$ for $c_i>0$ and $a_{i+1}=c_i$ and $b_{i+1}=b_i$ for $c_i<0$.