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:
$$\log(2^x) + \log(2^{-x}) = \log(10)$$
$$x \cdot \log(2) - x\cdot \log(2) = 1$$
$$0 = 1$$
 A: To make this clearer, make the substitution $u = 2^x$. Then your equation becomes
$$u + \frac 1 u = 10$$
Multiply through by $u$:
$$u^2 + 1 = 10u$$
Bring everything to one side:
$$u^2 - 10u + 1 = 0$$
You can use the quadratic formula to find roots. Keep in mind that $u = \text{stuff}$ is what you're finding. Thus, since $x = \log_2 (u)$, you will then take the $\log_2$ of the solutions of the quadratic to find $x$.

Also, to address your mistake:
$$\log_b(x+y) \neq \log_b(x) + \log_b(y)$$
When you took the log of both sides, you cannot apply it term-by-term, you have to apply it to the entire side of the equation. The property you might be thinking of is
$$\log_b(xy) = \log_b(x) + \log_b(y)$$
A: 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.
A: Welcome to math.stackexchange!
Hint:
$$2^x + 2^{-x} = 10$$
is equivalent to:
$$2^{x} + \dfrac{1}{2^{x}} = 10$$
and we can clear the fraction by multiplying on both sides by $2^{x}$.
