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Yesterday, my uncle asked me this question:

Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$.

How can we do this? Note that this is not a diophantine equation since $x \in \mathbb{R}$ if you are thinking about Fermat's Last Theorem.

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marked as duplicate by Parth Kohli, Pedro Tamaroff, Dominic Michaelis, Start wearing purple, Lord_Farin May 23 '13 at 17:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@AdiDani: Can you rephrase that please? –  Parth Kohli Jan 7 '13 at 13:15
@DumbCow I've just suggested an edit that does this. Typically $n$ is reserved for use representing natural numbers, and $x$ represents reals. –  apnorton Jan 7 '13 at 13:17
@anorton: Yes, I realized that. Thanks. –  Parth Kohli Jan 7 '13 at 13:17

4 Answers 4

up vote 15 down vote accepted

For all $x_j>x_i$ and $0<a<1$, $a^{x_i}>a^{x_j}$ .

Hence \begin{align} \left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 < \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 = 0 \end{align}

for all $x>2$. Hence, there is no solution for $x>2$.

Similarly \begin{align} \left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 > \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 =0 \end{align}

for all $x<2$.

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$$f(x) = \left(\dfrac{3}{5}\right)^x + \left(\dfrac{4}{5}\right)^x -1$$

$$f^ \prime(x) < 0\;\forall x \in \mathbb R\tag{1}$$

$f(2) =0$. If there are two zeros of $f(x)$, then by Rolle's theorem $f^\prime(x)$ will have a zero which is a contradiction to $(1)$.

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isn't Rolle's Theorem much stronger than what we need –  dirty derwin Jan 7 '13 at 13:21

Let $f(x)=5^x-4^x-3^x$. Then $f(2)=0$.

If $k>0$, then




If $k<0$, then




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One looks for roots of the function $f:x\mapsto a^x+1-b^x$ with $a=\frac34$ and $b=\frac54$.

  • Since $a\lt1$, the function $x\mapsto a^x$ is decreasing.
  • Since $b\gt1$, the function $x\mapsto b^x$ is increasing.
  • Hence the function $f$ is decreasing.
  • And $f(\pm\infty)=\mp\infty$.

As such, the function $f$ has exactly one root. Since $f(0)=1$ this root is positive.

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