What value of x satisfies (\log_2 x × \log_3 x)/(\log_2 x + \log_3 x) = 2?

2024 AMC 12B Problem 8 — Solution

2024 AMC 12B Problem 8

Below is the professionally curated solution for Problem 8 of the 2024 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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Concepts:logarithm

Difficulty rating: 1460

What value of $x$ satisfies

$\frac{\log_2 x \cdot \log_3 x}{\log_2 x + \log_3 x} = 2?$

$25$

$32$

$36$

$42$

$48$

Solution:

Dividing top and bottom by $\log_2 x \cdot \log_3 x,$ the left side becomes $\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} = \frac{1}{\log_x 2 + \log_x 3} = \frac{1}{\log_x 6}.$ So $\dfrac{1}{\log_x 6} = 2,$ meaning $\log_x 6 = \dfrac12,$ i.e. $x^{1/2} = 6.$

Therefore $x = 36.$

Thus, the correct answer is C.

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