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

2024 AMC 12B Problem 8

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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

Divide numerator and denominator by

\log_2 x \cdot \log_3 x

to turn the left side into

\dfrac{1}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x}}

Since

\dfrac{1}{\log_2 x} = \log_x 2

and

\dfrac{1}{\log_3 x} = \log_x 3,

the equation becomes

\log_x 6 = \dfrac12

Dividing top and bottom by

\log_2 x \cdot \log_3 x,

the left side becomes

\begin{gathered} \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}. \end{gathered}

\dfrac{1}{\log_x 6} = 2,

meaning

\log_x 6 = \dfrac12,

i.e.

x^{\frac{1}{2}} = 6.

Therefore

x = 36.

Thus, the correct answer is C.