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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?

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    $\begingroup$ Isn't it enough to divide both sides by $160^x$ and exploit convexity? $\endgroup$ – Jack D'Aurizio Jun 13 '17 at 19:49
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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.

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  • $\begingroup$ Is there any way to find the answer without testing? $\endgroup$ – Arnaldo Jun 13 '17 at 19:58
  • $\begingroup$ @Arnaldo Since $f$ is a decreasing function, our equation has one root maximum. But $3$ is a root, which gives the answer: $\{3\}$. It's not testing, it's just solution of your problem. $\endgroup$ – Michael Rozenberg Jun 13 '17 at 20:01
  • $\begingroup$ What I meant is once you know that you have just one solution, you then started to test which value of $x$ could give you the solution. Did you do that process? $\endgroup$ – Arnaldo Jun 13 '17 at 20:10
  • $\begingroup$ @Arnaldo The graph of a decreasing function can cross the $x$- axis in one point maximum. Draw it! We see that $(3,0)$ is a common, Thus, it's an unique point. $\endgroup$ – Michael Rozenberg Jun 13 '17 at 20:13
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    $\begingroup$ Thanks Michael, we are talking the same thing but just using different words. $\endgroup$ – Arnaldo Jun 13 '17 at 20:14
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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 $$\frac{2}{160^{1/3}} < \frac{3}{160^{1/3}} < \frac{5}{160^{1/3}} < 1 $$

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  • $\begingroup$ Is there any way to find the answer without testing? $\endgroup$ – Arnaldo Jun 13 '17 at 19:59
  • $\begingroup$ @Arnaldo: This answer proves there can only be one solution. So once we have found one, then we are sure to be done. $\endgroup$ – mathreadler Jun 13 '17 at 20:50
  • $\begingroup$ @mathreadler: You are right. My question was how to find such solution. After talk a little with Michael, I'm convinced that if we don't use a numerical method we must have luck to have a easy solution and find it by testing some values of $x$ or use some graph interpretation. $\endgroup$ – Arnaldo Jun 13 '17 at 21:01
  • $\begingroup$ If you want a closed-form solution, there will be very little hope unless $x$ is an integer. Since $160^x$ must be the cube of an integer, that integer must be divisible by $3$. So the first thing to try is $x=3$. $\endgroup$ – Robert Israel Jun 13 '17 at 21:20
  • $\begingroup$ Thank you Robert. You got my point. $\endgroup$ – Arnaldo Jun 13 '17 at 21:30

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