Solving the equation $2^x+2^{-x} = 10$ I'm trying to figure out how to solve this equation in terms of $x$.
Mistake I made so far is:
$$\log(2^x) + \log(2^{-x}) = \log(10)$$
$$x \cdot \log(2) - x\cdot \log(2) = 1$$
$$0 = 1$$

Let $2^x=y$
The equation becomes: $y^2-10y+1=0$. This quadratic equation has two solutions:
$$y = \frac{10 \pm \sqrt{100-4}}{2}  = 5 \pm 2\sqrt{6}.$$
So you get $x = \log_2(5+2\sqrt{6})$ and $x = \log_2(5-2\sqrt{6})$ as solutions.
