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.
 A: Let $f(x)=5^x-4^x-3^x$. Then $f(2)=0$.
If $k>0$, then
$f(2+k)=f(2+k)-f(2)$
$=25(5^k-1)-16(4^k-1)-9(3^k-1)$
$>25(5^k-1)-16(5^k-1)-9(5^k-1)=0$.
If $k<0$, then
$f(2+k)=f(2+k)-f(2)$
$=25(5^k-1)-16(4^k-1)-9(3^k-1)$
$<25(5^k-1)-16(5^k-1)-9(5^k-1)=0$.
A: 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.
A: $$f(x) = \left(\dfrac{3}{5}\right)^x + \left(\dfrac{4}{5}\right)^x -1$$
$$f^ \prime(x) < 0\;\forall x \in \mathbb R\tag{1}$$
$f(2) =0$. 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 $(1)$.
A: 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 $x<2$. 
