Solving an equation with exponentials

Question

$$2^x+4^x+12=0$$

How exactly am I supposed to solve this? Am I supposed to get $x$ alone or solve it another way?

Answer 1

For any $x \in \mathbb{R}$, $2^x > 0$ and $4^x > 0$, therefore

$$ 2^x + 4^x + 12 > 0 + 0 + 12 = 12 > 0 $$

Therefore there is no real solution.

Answer 2

let $u = 2^x$, then $4^x = (2^2)^x = 2^{2x} = 2^{2x} = (2^x)^2 = u^2$. Thus, $2^x + 4^x + 12 = 0$ becomes, $$u + u^2 + 12 = 0$$ Using the quadratic equation will solve $u$, which is really $2^x$. To solve for the $x$, just take $\log$ on both sides of the solution, then after rearranging, you should be able to solve for $x$.