I am attempting to solve the following equation:
$$ 2\log_2x - 1 = \log_2(x + 12)\\ 2\log_2x - \log_2(x + 12) = 1\\ \log_2 x^2 - \log_2(x + 12) = 1\\ \log_2 \left(\frac{x^2}{x + 12}\right) = 1\\ 2^1 = \frac{x^2}{x + 12}\\ 2(x + 12) = x^2\\ 2x + 24 = x^2\\ 0 = x^2 - 2x - 24\\ 0 = (x - 6)(x + 4) $$
Is $-4$ a valid solution?
$$ 2\log_2(-4) - 1 = \log_2(-4 + 12)\\ $$
$-4$ is not a valid solution, because the logaritmic function is not defined for negative numbers. In fact, suppose $-4$ is a valid solution and write $log_{2}(-4)=x\Rightarrow 2^{x}=-4$. That is a contradiction, because there is no $x\in\mathbb{R}$ such that $2^{x}<0$