13.45 Given $\log _{2} a+\log _{2} b \geqslant 6$, then the minimum value of $a+b$ is
(A) $2 \sqrt{6}$.
(B) 6 .
(C) $8 \sqrt{2}$.
(D) 16 .
(E) None of the above.
(20th American High School Mathematics Examination, 1969)
[Solution] From the given, we have $\log _{2} a+\log _{2} b=\log _{2} a b \geqslant 6$, then $a b \geqslant 2^{6}$. Therefore, $\quad a+b \geqslant 2 \sqrt{a b} \geqslant 2 \cdot 2^{3}=2^{4}$.
Thus,
$$
(a+b)_{\text {min }}=16 \text {. }
$$

So the answer is $(D)$.