exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now

$(\sqrt{(5+2\sqrt6)^x}+(\sqrt{(5-2\sqrt6)^x}=10$
We have, $5+2\sqrt6 = (\sqrt{2} + \sqrt{3})^2$ and $5 - 2\sqrt6 = (\sqrt3-\sqrt2)^2$
The equation $\Leftrightarrow \left(\sqrt{2} + \sqrt{3}\right)^x + \left(\sqrt3-\sqrt2\right)^x = 10$
Let $$t = \left(\sqrt3+\sqrt2\right)^x = \frac{1}{\left(\sqrt3-\sqrt2\right)^x}$$ with $t\neq 0$
The equation $\Leftrightarrow t + \frac1t - 10 = 0 \Leftrightarrow t^2 - 10t + 1 = 0 $
$\Leftrightarrow t = 5 \pm 2\sqrt6$
With case $t = 5 + 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 + 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5+2\sqrt6) = 2$
With case $t = 5 - 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 - 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5-2\sqrt6) = -2$.
Done!
