If it is known that $\log_2(a)+\log_2(b) \ge 6$, then the least value that can be taken on by $a+b$ is:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$
We use the logarithm property of addition:
\begin{align*} \log_2(a)+\log_2(b) \ge 6 &= \log_2(ab) \ge 6\\ &\Rightarrow 2^{log_2(ab)} \ge 2^6\\ &= ab \ge 64 \end{align*}
Due to the Quadratic [Optimization](https://artofproblemsolving.com/wiki/index.php/Optimization) or the [AM-GM Inequality](https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality), the least value is obtained when $a = b$.
Therefore, $a = b = 8 \Rightarrow a + b = \boxed{(D)16}$