Is there really only one answer?

${2}^{x}+{3}^{x}={5}^{x}$ Substitute $x=1\ \implies 2 + 3=5$ ,satisfying the equation.It can be concluded that $x=1$ is a root.But is it the ONLY possible answer?That's the real question.Divide both sides by ${5}^{x}$ to get ${\left(\dfrac{2}{5}\right)}^{x}+{\left(\dfrac{3}{5}\right)}^{x}=1$ .Now we have to cases:

Case # 1: x<1

If $x<1$ ,then: $\begin{cases} {\left(\dfrac{2}{5}\right)}^{x}>\dfrac{2}{5}\\ {\left(\dfrac{3}{5}\right)}^{x}>\dfrac{3}{5}\end{cases} \\ \implies {\left(\dfrac{2}{5}\right)}^{x}+{\left( \dfrac {3}{5}\right)}^{x}>1$ Case # 2:x>1

If $x>1$ ,then: $\begin{cases}{\left(\dfrac{2}{5}\right)}^{x}<\dfrac{2}{5}\\ {\left(\dfrac{3}{5}\right)}^{x}<\dfrac{3}{5}\end{cases}\\ \implies {\left(\dfrac{2}{5}\right)}^{x}+{\left(\dfrac{3}{5}\right)}^{x}<1$ Therefore,it can be observed that neither $x<1$ nor $x>1$ satisfies the equation.Thus, $\boxed{x=1}$ is the only root of the equation ${2}^{x}+{3}^{x}={5}^{x}$ .

Because $5^{2}\neq2^{2}+3^{2}$ , we know there is only one solution (x = 1) because of Fermat's theorem.

Fermat's? Correct me if I'm wrong.

All you have to do is turn it into a fraction and add it. 2 and 3 will become the numerator and 5 will be the denominator.

So it's 3/5+2/5= 5/5 which is equal to 1