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

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 $x<2$. 
