Plug in the Values, or should you?

$2^x + 3^x + 4^x - 5^x = 0$

$( \frac{2}{5})^{x} + (\frac{3}{5})^{x} + (\frac{4}{5})^{x} -1 = 0$

Let this be $f(x)$ , differentiating it:-

$f'(x) < 0$ Thus, it is a strictly decreasing function.

Now, at x=0, f(x) = 2

at $x= \infty , f(x) = -1$

at $x = -\infty , f(x) = \infty$

Thus it cuts $x$ axis at only one point between x=0 and x= $\infty$

Using the intermediate value theorem since the expression is continuous when graphed, the LHS is $4$ when $x=2$ and the LHS is $-26$ when $x=3$ . Since it has to cross the $x$ -axis, there is a root between $2$ and $3$ . After trying some other values, it can be concluded that the graph never touches the $x$ -axis. Thus, there is $1$ real root.

Very near x=2.373294 only satisfies the equation. One solution.