Finding all real solutions to the equation $3^x+4^x=5^x$

Question

Find all real solutions to the equation $$3^x+4^x=5^x.$$

My attempt: It is evident that $x=2$ is a solution.
However, I think that there are no other solutions. So, I define a function $f(x)=3^x+4^x-5^x$.
Differentiating w.r.t $x$, we get $$f'(x)=3^x\ln 3+4^x\ln 4-5^x\ln 5,$$ but that doesn't take me anywhere. Please help. Thank you.

score 7 · Accepted Answer · answered Mar 04 '15 at 10:48

7

Hint: Divide through by $5^x$ to produce the equivalent equation $$\left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1.$$ The l.h.s. is a strictly decreasing function of $x$.

answered Mar 04 '15 at 10:48

Travis Willse

99,363

And convert to $\sin \theta$ and $\cos \theta$ would be a good idea you think? – Swadhin Mar 04 '15 at 10:55
I don't see immediately how that would help. The key observation now is that (by definition) a strictly decreasing function achieves any given value at most once. – Travis Willse Mar 04 '15 at 10:56
Yes, thank you. – Swadhin Mar 04 '15 at 10:58
You're welcome, I'm glad you found it useful. – Travis Willse Mar 04 '15 at 11:18

Finding all real solutions to the equation $3^x+4^x=5^x$

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