$3^x+4^x+5^x=6^x$ has exactly one real solution

Question

$3^x+4^x+5^x=6^x$ has exactly one real solution.

I took $f(x)=\frac{3^x} {6^x} + \frac{4^x}{ 6^x}+ \frac{5^x}{6^x} -1 $ it's decreasing function since $f\prime(x) < 0$. But how can I say it has exactly one real root? Please help me. Thanks in advance.

Answer 1

If the derivative is always negative, this means the function is decreasing. Notice $$\lim_{x \to +\infty} f(x) = -1 \quad \text{and} \quad \lim_{x \to -\infty} f(x) = +\infty,$$ and by continuity there must be a root. The function is decreasing implies that there can be only one root. Indeed, if $x_1 < x_2$ were both roots then $$0 = f(x_1) > f(x_2) = 0$$ and you conclude $0> 0$.

Answer 2

Since the function $f(x)$ is strictly decreasing and differentiable on $\mathbb R$ , it cannot have two distinct real roots because the theorem of Rolle would guarantee a point $c$ with $f'(c)=0$ which contradicts your result. With the comment of "Integrand", you can show that there is actually a real root, hence exactly one.