1969 AHSME Problems/Problem 25: Difference between revisions

Revision as of 04:53, 18 August 2016

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, $a = b$ . Therefore, $a = b = 8 \Rightarrow a + b = \boxed{D}$

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{D}</math>
+We use the logarithm property of addition:
+<cmath>
+\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*}</cmath>
+Due to the Quadratic [[Optimization]], <math>a = b</math>.
+Therefore, <math>a = b = 8 \Rightarrow a + b = \boxed{D}</math>
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1969 AHSME Problems/Problem 25: Difference between revisions

Revision as of 04:53, 18 August 2016

Problem

Solution

See also