# Showing that a equation has exactly on real root

I have a homework in real analysis and I'm very confused about it. I would be very thankful, if you could give me any ideas/tips or solutions how to get this task done. The task is as follows:

Prove that equation $$3^{x} + 4^x = 5^x$$ has exactly one real root. Hint that has been given: examine the function $f(x) = \frac{3^x}{5^x} + \frac{4^x}{5^x} - 1$.

• So $f(2)=0$. Now what can you say about the behavior of $f(x)$ around 2? – Laars Helenius Oct 27 '17 at 6:22

## 2 Answers

$$f'(x)=(3/5)^x \ln \frac35+(4/5)^x \ln \frac45;$$ $$\ln \frac35<0,\;\;\ln \frac45<0$$ $$\forall x\in\mathbb R\quad (3/5)^x>0,\;\;(4/5)^x>0$$ Therefore $\forall x\in\mathbb R\;\; f'(x)<0$, thus $f(x)$ is strictly decreasing $\Rightarrow$ $f(x)=0$ has only one root $x=2$ (because for $x>2\;$ $f(x)<f(2)=0$ and for $x<2\;$ $f(x)>f(2)=0$).

It is clear that $x=2$ is a solution.You have to show that there are no other solutions.The following solution uses only elementary inequality method(without calculus).

Note that $(3/5)^2+(4/5)^2=1$.Again for any given $x>2$ we have $(3/5)^x<(3/5)^2 ,(4/5)^x<(4/5)^2$ then $3^x+4^x<5^x$ for any given $x>2$.Now try yourself to prove similar useful inequality for $x<2$.