I would like to prove that the equation $ 3^x+4^x=5^x $ has only one real solution ($x=2$)

I tried to study the function $ f(x)=5^x-4^x-3^x $ (in order to use the intermediate value theorem) but I am not able to find the sign of $ f'(x)= \ln(5)\times5^x-\ln(4)\times4^x-\ln(3)\times3^x $ and I can't see any other method to solve this exercise...

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    $\begingroup$ Maybe it would be easier to write the equation as $\left(\frac 35\right)^x +\left(\frac 45\right)^x=1$. $\endgroup$ – Davide Giraudo Sep 4 '11 at 13:51
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    $\begingroup$ Write $x = 2n + r$ for $n\in\mathbb{Z}$ and $r\in[0,2).$ Then if $x>2,$ $$5^x = (4^2 + 3^2)^n 5^r = 4^{2n}5^r + 3^{2n}5^r + 5^r\displaystyle\sum_{i = 1}^{n-1} \binom {n} {i} 4^{2i}3^{2(n-1)} > 4^{2n + r} + 3^{2n+ r} = 4^x + 3^x.$$ $\endgroup$ – jspecter Sep 4 '11 at 13:56
  • $\begingroup$ And if $x \in \mathbb{Q}_{<0}$ the lhs lies in $\mathbb{Z}_{(5)}$ whereas the rhs does not. $\endgroup$ – jspecter Sep 4 '11 at 14:03
  • $\begingroup$ @Davide, you seem to have hit it over the head with that. Post it as an answer! $\endgroup$ – Ragib Zaman Sep 4 '11 at 14:06
  • $\begingroup$ @Ragib: now it's too late, the answer will be the same as Beni Bogosel's proof. $\endgroup$ – Davide Giraudo Sep 4 '11 at 14:48

One direct method is to divide directly by $5^x$ and get $1=(3/5)^x+(4/5)^x$. From here it is clear that the RHS is strictly decreasing, and there is a unique solution. Almost all exponential equations can be treated this way, by transforming them to

  • one increasing function equal to one decreasing function

  • one increasing/decreasing function equal to a constant.

  • $\begingroup$ Thanks for sharing this method. I am wondering how could we prove that (a) the function is continuously decreasing (other than by piloting or inspection) and (b) there is a unique solution. Thanks. $\endgroup$ – NoChance Jul 2 '19 at 15:33
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    $\begingroup$ I guess it is possible to prove that the function is strictly increasing without calculus, just using the basic properties of the exponential function $x \mapsto a^x$ for $a>1$ or $a<1$. This will prove that there is at most one solution. The uniqueness comes from the fact that in this case we can identify the solution $x=2$. $\endgroup$ – Beni Bogosel Jul 2 '19 at 20:56
  • $\begingroup$ Thank you for your answer. $\endgroup$ – NoChance Jul 2 '19 at 20:58

If we insert the known solution we can write $$ 5^{2+x} = 4^{2+x} + 3^{2+x} $$ asking, whether there might another solution exist besides $x=0 $ . Then we can rewrite, putting the $5^x$ to the rhs:
$$ 5^2 = 4^2\cdot 0.8^x + 3^2\cdot 0.6^x $$ Then if the exponents $x$ on the rhs are zero, we have the known solution. But if $x$ increases over zero, then the values of both summands decrease simultaneously, thus the equality can no more hold.
The analogue occurs for decreasing $x$: both summands increase over their squares simultaneously, so there is no other solution possible. QED.

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    $\begingroup$ Best possible answer i think. $\endgroup$ – mick Sep 29 '12 at 19:18
  • $\begingroup$ +1 Marvelous answer...but kind of related to Beni's answer too $\endgroup$ – imranfat Mar 2 '17 at 4:10

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