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

Solution 1:

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

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

$f(2) =0$. 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 $(1)$.

Solution 2:

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 $x<2$.