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This question already has an answer here:

Let define $f:\mathbb{R}\rightarrow\mathbb{R}$ as $f(x)=3^x+4^x-5^x$.

Prove that there exists only one $x_0$ such as $f(x_0)=0$.

My approach:

We can see that $\lim_{x\rightarrow-\infty}f(x)=0$ and $f(2)=0$. I don't know how to use the derivatives of $f$ to prove that it is firstly increasing and then decreasing for $x\in(-\infty,2)$ and for $x\in(2,\infty)$ it is only decreasing.

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marked as duplicate by Dietrich Burde, Thomas Shelby, Javi, José Carlos Santos real-analysis Mar 20 at 16:16

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.

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For $x>2$ we have $3^{x}+4^{x}=(9)3^{x-2}+(16)4^{x-2}<(9)5^{x-2}+(16)5^{x-2}=5^{x}$ and the inequalities gets reversed for $x<2$. Hence $x=2$ is the only solution.

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Easier way to see it is $$ 3^x+4^x=5^x \qquad\Longleftrightarrow\qquad f(x)=\left(\frac{3}{5}\right)^x +\left(\frac{4}{5}\right)^x-1=0 $$ Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.

Also $f(2)=0$, and hence $x=2$ is the one and only root of $f$.

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