Solving radical questions

Question

I came across this question in Maths olympiad and am stuck as to how to solve it.

$(\sqrt 2)^x + (\sqrt 3)^x = (\sqrt{13})^x$, find $x$.

Answer 1

We can square both sides, then rearrange and square again to remove the radicals to get a slightly easier form but there doesn't appear to be a closed form for the solution and WolframAlpha can't find one either.

$$\begin{align} (\sqrt{2})^x+(\sqrt{3})^x&=(\sqrt{13})^x\\ 2^x+2\cdot(\sqrt{2}\sqrt{3})^x+3^x&=13^x\\ 4\cdot6^x&=13^{2x}-2\cdot13^x(2^x+3^x) +(2^x+3^x)^2\\ 4\cdot6^x+2\cdot13^x(2^x+3^x)&=13^{2x} +2^{2x}+2\cdot6^x+3^{2x}\\ 2&=\frac{13^{2x} +2^{2x}+3^{2x}}{6^x+26^x+39^x}\\ \end{align}$$

If we allow numerical solutions in place of analytical ones, we may define $f(x)=(\sqrt{2})^x+(\sqrt{3})^x-(\sqrt{13})^x$, so we can find a sequence converging to the root with the Newton-Raphson method. It's more convenient to use the original formula for this, than to use the one with no radicals since the algebra is much simpler.

$$\begin{align}f'(x)&=(\sqrt{2})^x\ln\sqrt2+(\sqrt{3})^x\ln\sqrt3-(\sqrt{13})^x\ln\sqrt{13}\\ x_{n+1}&=x_n-\frac{f(x_n)}{f'(x_n)}\\ &=x_n-2\frac{2^{x_n/2}+3^{x_n/2}-13^{x_n/2}}{2^{x_n/2}\ln2+3^{x_n/2}\ln3-13^{x_n/2}\ln{13}}\\ \end{align}$$

If we adapt this into the python3 code below, and set $x_0=0.8$, we find that $x\approx0.834\,864\,863\,164\,620$. However this solution doesn't have any obvious possible analytic forms.

import math as m

def x(n):
    if n==0:return 0.8
    else: return x(n-1)-2*((2**(x(n-1)/2)+3**(x(n-1)/2)-13**(x(n-1)/2))/(2**(x(n-1)/2)*m.log(2)+3**(x(n-1)/2)*m.log(3)-13**(x(n-1)/2)*m.log(13)))

for n in range(0,6):print(n,x(n))