2020 AMC 12B Problem 13

Attempt Problem 13 of the 2020 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

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13.

Which of the following is the value of log26+log36?\sqrt{\log_2 6 + \log_3 6}?

11

log56\sqrt{\log_5 6}

22

log23+log32\sqrt{\log_2 3} + \sqrt{\log_3 2}

log26+log36\sqrt{\log_2 6} + \sqrt{\log_3 6}

Answer: D
Concepts:logarithmalgebraic manipulation
Difficulty rating: 1590
Solution:

Let a=log23,a = \log_2 3, so log32=1a.\log_3 2 = \tfrac1a. Then log26+log36=(1+log23)+(1+log32)=2+a+1a. \begin{gathered} \log_2 6 + \log_3 6 \\ {}= (1 + \log_2 3) \\ \quad {}+ (1 + \log_3 2) \\ {}= 2 + a + \frac1a. \end{gathered}

Meanwhile (log23+log32)2\left(\sqrt{\log_2 3} + \sqrt{\log_3 2}\right)^2 =a+1a+2a1a= a + \frac1a + 2\sqrt{a \cdot \tfrac1a} =a+1a+2,= a + \frac1a + 2, which equals the expression above.

Taking square roots, log26+log36\sqrt{\log_2 6 + \log_3 6} =log23+log32.= \sqrt{\log_2 3} + \sqrt{\log_3 2}.

Thus, the correct answer is D.

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