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Recently I encountered an exponential equation in form of $a^x + x^a = b^x$, where both $a,b$ are integers. The tricky part is $x$ appears in both the base and the exponential part.

Is this kind of equation solvable?

For instance, $6^x + x^6 = 10^x$. Where should I start from?

EDIT: There is a similar question. But this one differs from the other one in a way that it has two constants $a, b$

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    $\begingroup$ Start from $x=0$. Is $x$ also an integer? For real $x$ it is more a less a duplicate of this question. $\endgroup$ – Dietrich Burde Jun 4 '18 at 18:49
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I would start by plotting $y=10^x-6^x-x^6$ on Wolfram Alpha to get an idea of the answer. We easily check that $x=0$ and $x=2$ are solutions. The solution between $2$ and $3$ can be computed numerically.

If you want to prove that these are the only solutions, that is another matter.

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  • $\begingroup$ What is amazing is that using a Taylor expansion built at $x=\frac52$, the roots of the quadratic are $1.99026$ and $2.80763$ while the solution for the largest root is $2.81205$. $\endgroup$ – Claude Leibovici Jun 6 '18 at 9:07
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As far as proving $x=0$ and $x=2$ are the only solutions: If we are assuming real integers, we see that

  1. $|10^x - 6^x|$ is strictly less than 1 for $x \leq -2$ on the one hand, so $x$ has to be at least $-1$.

  2. $10^x$ grows much more rapidly than $x^6+6^x$ as $x$ gets large, in fact, by for all $x \ge 6$, the quantity $6^x+x^6$ is less than $10^x$, and $6^(x+1) + (x+1)^6 \leq 9 \times (6^x+x^6)$, whereas $10^{x+1} = 10 \times 10^x$. Thus the strict inequality $6^x+x^6 < 10^x$ holds for all $x \ge 6$. So on the other hand $x \leq 5$.

Thus all integer solutions are in the set $\{-1,0,1,2,3,4,5\}$. This can be checked exhaustively (if nothing else)

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