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I have been trying to solve the following equation:

$5^x+7^x=12^x.$

Obviously, x=1 is a solution but how do I prove that there are no other solutions.

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  • $\begingroup$ Is $x$ supposed to be an integer, or is it a real number? Is it allowed to be negative? $\endgroup$ Sep 29, 2014 at 14:23
  • $\begingroup$ if you're familiar with derivatives you could just use that $\endgroup$
    – mm-aops
    Sep 29, 2014 at 14:23
  • $\begingroup$ X is real number. No, I cannot use derivatives in the solution. $\endgroup$
    – chen h.
    Sep 29, 2014 at 14:34
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    $\begingroup$ For $x > 2$, you can use Fermat's last theorem! $\endgroup$
    – rlms
    Sep 29, 2014 at 20:33
  • $\begingroup$ FLT doesn't work for all $x \in \mathbb{R}$. $\endgroup$
    – Confuse
    Feb 7, 2017 at 7:43

1 Answer 1

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we'll show it has no solutions for $x > 1$, hope you can use the idea to deal with the other case. your observation (that $1$ is a solution) will be crucial. write $x = 1 + y$ with $y > 0$, then $$ 5^x + 7^x = 5 \cdot 5^y + 7 \cdot 7^y < (5+7)\cdot 7^y < 12 \cdot 12^y = 12^x$$ inequalities are strict because $y > 0$.

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