Solve $(2^x+3^x+5^x)^3=160^x$

Question

The question is to find all real solutions of the equation:

$$(2^x+3^x+5^x)^3=160^x$$

Using Wolfram, I can check that $x=3$ is the only solution, but I'm having trouble trying to find it by hand.

First idea was try to use AM-GM inequality. Some algebric manipulation also didn't work.

Any idea?

Answer 1

It's $f(x)=0$, where $$f(x)=\left(\frac{2}{\sqrt[3]{160}}\right)^x+\left(\frac{3}{\sqrt[3]{160}}\right)^x+\left(\frac{5}{\sqrt[3]{160}}\right)^x-1.$$ But $f$ is a decreasing function and the rest for you.

Answer 2

The equation can be written as

$$ \left(\frac{2}{160^{1/3}}\right)^x + \left(\frac{3}{160^{1/3}}\right)^x + \left(\frac{5}{160^{1/3}}\right)^x = 1$$

The left side is a decreasing function of $x$, since $$\frac{2}{160^{1/3}} < \frac{3}{160^{1/3}} < \frac{5}{160^{1/3}} < 1 $$