solve for $x$, $(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a$

Question

Find the value of $x$ when $$(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a.$$See, by hit and trial method it is clear that $x=2$ is a solution. But I failed to solve this explicitly to get the solutions.
My Attempt: \begin{align*} &(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a \\ \implies \> & (a+ \sqrt{a^2-1})^{x/2}+(a- \sqrt{a^2-1})^{x/2}=2a\\ \implies \>& (a+ \sqrt{a^2-1})^x+(a- \sqrt{a^2-1})^x+2(a+ \sqrt{a^2-1})^{x/2}(a- \sqrt{a^2-1})^{x/2} = 4a^2.\end{align*} I have no idea how to proceed after this. Also I tried by multiplying conjugate up and down, but I failed. Please help me to solve this. Thanks in advance.

Answer 1

put $u=(\sqrt{a+ \sqrt{a^2-1}})^x$

then taking conjugate,$\frac{1}{u}$=$(\sqrt{a- \sqrt{a^2-1}})^x$ or we have to solve

$u+\frac{1}{u}=2a$ which can be easily solved.

Answer 2

Hint : Let $y=a+\sqrt{a^2-1}$. The equation you have to solve is equivalent to $$y^{x/2} + \frac{1}{y^{x/2}} = y +\frac{1}{y}$$