1969 AHSME Problems/Problem 25

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Problem

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:

$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$

Solution

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 or the AM-GM Inequality, the least value is obtained when $a = b$. Therefore, $a = b = 8 \Rightarrow a + b = \boxed{(D)16}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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