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This problem has been distracting me for quite some time. I have found some values for x with different values of a and b using brute force, but could not find any patterns in the results. I was wondering if there was a better way to solve for x.

Here is one solution I found: if $x = 2.446481944$ (approx.) then $4^x + 5^x = (4 + 5)^2 = 81$

Obinna Nwakwue
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mPrime
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  • Are we allowed to work mod 2? :P – TomGrubb Jun 12 '17 at 01:32
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    Your example suggests you have in mind $a,b$ are natural numbers. It would improve your Question to state any such restrictions on $a,b$ that are tacitly assumed (e.g. are they positive reals? etc.). – hardmath Jun 12 '17 at 01:36
  • The values of a and b are positive whole integers. – mPrime Jun 12 '17 at 01:38
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    The function $f(x)=a^x+b^x$ is continuous, increasing and unbounded on $\mathbb{R}^+$, so it will take all values $\ge f(0) =2,$ exactly once. Since $(a+b)^2 \ge 4$ the equation will therefore have a unique solution, which can be determined numerically, but there is no nice closed form in general. – dxiv Jun 12 '17 at 01:45
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    @dxiv Unless $a=b=1$, in which case there is no solution. – Robert Israel Jun 12 '17 at 03:50
  • @RobertIsrael Indeed. (For values of $1$ such that $1+1 \ne 0,$ ;-)). – dxiv Jun 12 '17 at 03:56
  • There are some mathematical ways to do that, but no analytical way. – Ghartal Jun 12 '17 at 07:52

1 Answers1

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As Robert Israel commented, there is no explicit solution for $x$ except in a few cases such as $a=b^n$ $(n=0,1,2,3,4)$ where the problem reduces to polynomials.

So, for the most general case, you need some numerical method and the simplest is probably Newton considering that we look for the zero of equation $$f(x)=a^x+b^x-(a+b)^2=e^{x\log(a)}+e^{x\log(b)}-(a+b)^2$$ Without loss of generality, let us assume $a>b$ and rewrite $$e^{x\log(a)}=(a+b)^2-e^{x\log(b)}$$ and search for the zero of equation $$g(x)={x\log(a)}-\log\left((a+b)^2-e^{x\log(b)} \right)$$ which is better conditioned. You can start iterating using $$x_0=2\frac{\log(a+b)} {\log(a)}$$ Using your numbers $a=5$ and $b=4$, Newton method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.730424778 \\ 1 & 2.489833783 \\ 2 & 2.447205870 \\ 3 & 2.446482135 \\ 4 & 2.446481944 \end{array} \right)$$ which is the solution for ten significant figures.

If you plot function $g(x)$, you will notice that it is almost a straight line; this explains why the convergence is so fast.

Claude Leibovici
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