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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Is $x$ supposed to be an integer, or is it a real number? Is it allowed to be negative? – Mark Bennet Sep 29 '14 at 14:23
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if you're familiar with derivatives you could just use that – mm-aops Sep 29 '14 at 14:23
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X is real number. No, I cannot use derivatives in the solution. – chen h. Sep 29 '14 at 14:34
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2For $x > 2$, you can use Fermat's last theorem! – rlms Sep 29 '14 at 20:33
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FLT doesn't work for all $x \in \mathbb{R}$. – Confuse Feb 07 '17 at 07:43
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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 $y > 0$.
mm-aops
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