Which of the following is the value of √\log_2 6 + \log_3 6?

2020 AMC 12B Problem 13 — Solution

2020 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2020 AMC 12B, 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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Concepts:logarithm algebraic manipulation

Difficulty rating: 1590

13.

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

$1$

$\sqrt{\log_5 6}$

$2$

$\sqrt{\log_2 3} + \sqrt{\log_3 2}$

$\sqrt{\log_2 6} + \sqrt{\log_3 6}$

Solution:

Let $a = \log_2 3,$ so $\log_3 2 = \tfrac1a.$ Then $\log_2 6 + \log_3 6 = (1 + \log_2 3) + (1 + \log_3 2) = 2 + a + \frac1a.$

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

Taking square roots, $\sqrt{\log_2 6 + \log_3 6} = \sqrt{\log_2 3} + \sqrt{\log_3 2}.$

Thus, the correct answer is D.

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