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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 ?

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Can you see at least one solution for this? – Praphulla Koushik Jun 4 '14 at 14:07
@PraphullaKoushik $x=4$ ? But this is by manual counting. Is there a standard way of solving ? – Gaurang Tandon Jun 4 '14 at 14:07
I can not see any other solution than $x=4$.. I do not have any formal proof as of now.... – Praphulla Koushik Jun 4 '14 at 14:12
@PraphullaKoushik No, you must have a most remarkable proof of this which the comment is too small to contain :P – M. Vinay Jun 4 '14 at 15:19
@M.Vinay : you call that a sense of humor? Fine... :P – Praphulla Koushik Jun 5 '14 at 7:14
up vote 14 down vote accepted

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.

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This is indeed a very easy solution; but could you please tell why you use $x=4y$, and not $x=9y$ or $10y$? Would using different values affect the answer ? – Gaurang Tandon Jun 4 '14 at 14:23
That's just for simplification of notation, on the right hand side, we start with $(\sqrt{13})^{\large\frac{x}{2}} = 13^{\large\frac{x}{4}}$, and it's simpler to write $y$ than $\frac{x}{4}$. – Daniel Fischer Jun 4 '14 at 14:25
I see your idea. Definitely simple. – Gaurang Tandon Jun 4 '14 at 14:26

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:

  1. The function is strictly increasing for $x<2$
  2. $\displaystyle \lim_{x\to-\infty}f(x) = 0$
  3. The function is concave on $[2,3]$
  4. The function is decreasing for $x>3$
  5. $\displaystyle \lim_{x\to\infty}f(x) = -\infty$

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 $[3,\infty)$.

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This should definitely be a nice solution, but I am sorry that I do not understand it since I yet know not the limits and concaves and all. Thanks though :) – Gaurang Tandon Jun 4 '14 at 14:24

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 $f(x)<0$


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This solution seems as if we already know the answer is $4$, right? – Gaurang Tandon Jun 4 '14 at 14:22
As i know one solution I am checking if there are any on its right side (in real line) and if there are any on its left.... – Praphulla Koushik Jun 4 '14 at 14:24
Oh. I see. But again that thing: "As i know one solution" - but that solution is obtained by manual checking :( – Gaurang Tandon Jun 4 '14 at 14:26
I guess that one solution is very obvious to see... With all due respect to other answers this is not the best way to see this for sure... – Praphulla Koushik Jun 4 '14 at 14:28

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