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The problem is to prove for the function: $$ f(x) = 3^x+4^x-5^x$$ has only one root.

I know it isn't a hard problem, but I am really stuck on it so I would appreciate your help. I have calculated the derivative and calculated it to zero:

$$ f'(x)=3^x\ln(3)+4^x\ln(4)-5^x\ln(5)=0.$$ Dividing by $5^x\ln(5)$, we have $$\frac{3}{5}^x\ln(\frac{3}{5}) + \frac{4}{5}^x\ln(\frac{4}{5}) = 1 $$

Then I see no clear continuation.

Arctic Char
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Funny
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    $ln(a) /ln(b) \neq \ln(a/b) $ – kingW3 Oct 11 '20 at 10:27
  • The problem is to prove what for the function? That only one root ie solution for $f(x)=0$ exists? Then why are you setting $f‘(x)=0$? – Jonas Linssen Oct 11 '20 at 10:33
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    Look at $g(x)=(3/5)^x+(4/5)^x-1$ if $f(x_1)=0$ then $g(x_1)=0$. Prove that both $(3/5)^x,(4/5)^x$ are decreasing hence $g$ is decreasing. – kingW3 Oct 11 '20 at 10:35
  • The only solution is $x=2$. Because $f(1)=1$ and $f(3)=-34$ so for the intermediate value theorem $f(x)$ must have at least a root in $(1,3)$. By exploration we can easily see that $x=2$. Studying the properties of the derivatives and the limits at positive and negative infinity we see that $x=2$ is the unique root. – Raffaele Oct 11 '20 at 15:28

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We have $$f(x)=0\iff 3^x \left(1+\left(\frac43\right)^x-\left(\frac53\right)^x\right)=0\iff\left(\frac43\right)^x-\left(\frac53\right)^x=-1.$$

Can you finish it now? (Hint: The LHS in the last equation is decreasing.)

Maximilian Janisch
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