3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is $x$ supposed to be an integer, or is it a real number? Is it allowed to be negative? $\endgroup$ – Mark Bennet Sep 29 '14 at 14:23
  • $\begingroup$ if you're familiar with derivatives you could just use that $\endgroup$ – mm-aops Sep 29 '14 at 14:23
  • $\begingroup$ X is real number. No, I cannot use derivatives in the solution. $\endgroup$ – chen h. Sep 29 '14 at 14:34
  • 2
    $\begingroup$ For $x > 2$, you can use Fermat's last theorem! $\endgroup$ – rlms Sep 29 '14 at 20:33
  • $\begingroup$ FLT doesn't work for all $x \in \mathbb{R}$. $\endgroup$ – Confuse Feb 7 '17 at 7:43
10
$\begingroup$

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 $y > 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.