How do I solve for $x$ in the logarithms $10^x+10^{-x}=4$

Question

$x$ should equal to.. I don't know how to get to the answer.

$x= \log_{10} (2\pm \sqrt{3})$

Thank you.

Similarly: $7^x+7^{-x}=4$. The answer is the same.

Answer 1

Hint : Use $a^{-x}=\frac{1}{a^x}$ and substitute $t=a^x$ to get a quadratic equation