Find the number of values of $x$ in $(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}$. Question:

Find the number of values of $x$ with $$(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}.$$

My attempt:
I tried setting $p = \sqrt{2}$, $q = \sqrt{3}$, and resetting the equation to:
$$p^x + q^x = \sqrt{p^4+q^4}^{\frac{x}{2}}$$
But I don't still think that expansion would probably solve it. Anyone has a better way ?
 A: This is how I would attack your problem:
take a look at the function $$f(x) = \sqrt2^x + \sqrt3^x - \sqrt{13}^{\frac x2}.$$
If you plot this function, you can get the idea that it may only have one root: Plot of the function.
Now, by no means is this a strict proof, but it does help you to figure out the proof by looking at some of the properties of the function which can be proven:


*

*The function is strictly increasing for $x<2$

*$\displaystyle \lim_{x\to-\infty}f(x) = 0$

*The function is concave on $[2,3]$

*The function is decreasing for $x>3$

*$\displaystyle \lim_{x\to\infty}f(x) = -\infty$


Now, from 1. and 2. you can conclude that the function has no roots on $[-\infty, 2]$. From 3., you can conclude that it also has no roots on $[2,3]$. From 4. and 5., you can conclude that the function has only one root on $[3,\infty)$.
A: Divide by $(\sqrt{13})^{\large\frac{x}{2}}$, and write $x = 4y$. You obtain
$$\left(\frac{4}{13}\right)^y + \left(\frac{9}{13}\right)^y = 1.$$
For $0 < a < 1$, the sole solution of $a^y + (1-a)^y = 1$ is $y = 1$, since $y\mapsto a^y$ and $y\mapsto (1-a)^y$ are both strictly decreasing.
A: For $f(x)=(\sqrt{2})^x+(\sqrt{3})^x-(\sqrt{13})^{\frac{x}{2}}$


*

*for $x>4$ we have $f(x)>0$

*for $x<4$ we have $f(x)<0$


So???
