If $\left(\sqrt2\right)^x + \left(\sqrt3\right)^x = \left(\sqrt{13}\right)^{\frac{x}{2}}$ then the number of values of $x$ is?

Question

If $\left(\sqrt2\right)^x + \left(\sqrt3\right)^x = \left(\sqrt{13}\right)^{\frac{x}{2}}$ then the number of values of $x$ is?

It is an exponent topic question tried squaring method but couldn't get to the right answer. The answer is $1$.

Answer 1

We have $$(4/13)^{x/4}+(9/13)^{x/14}=1$$

Now clearly $x/4=1$ is a solution

As the left hand side is monotonic decreasing, there can be no more solution

This approach can be applied for

$$(\cos^2t)^n+(\sin^2t)^n=1$$

Answer 2

Just for your curiosity (but too long for a comment).

For your specific problem, as @lab bhattacharjee answered, the problem is "quite simple" because $2^2+3^2=13$.

Making the problem more general, suppose that you want $x$ such that $$\left(\sqrt 2\right)^x+\left(\sqrt 3\right)^x=\left(\sqrt k\right)^{\frac x 2}\implies 2^{x/2}+3^{x/2}=k^{x/4}$$ This means that you are looking for the zero of function $$f(x)=k^{x/4}-\big[ 2^{x/2}+3^{x/2}\big]$$ which is not very pleasant (have a look at its plot). However, if you consider the function $$g(x)=\frac x 4 \log(k)-\log\big[ 2^{x/2}+3^{x/2}\big]$$ it is almost the problem of the intersection of two straight lines. Using Taylor expansion built around $x=0$, you will have $$g(x)=\log (2)+\frac{1}{4} x (\log (3)+\log (2)-\log (k))+O\left(x^2\right)$$ which gives as an estimate $$x_0=\frac {4 \,\log(2)}{\log \left(\frac{k}{6}\right)}$$ For $k=13$, this gives $x_0 \sim 3.58591$ This is in fact the result of the first iteration of Newton method.

Performing a second iteration would give $x_1 \sim 3.99515$