2024 AMC 12B Problems/Problem 8

Problem

What value of satisfies

Solution 1

We have \begin{align*} &\log_2x\cdot\log_3x=2(\log_2x+\log_3x) \\ &1=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\log_3x} \\ &1=2(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\ &1=2(\log_x3+\log_x2) \\ &\log_x6=\frac{1}{2} \\ &x^{\frac{1}{2}}=6 \\ &x=36 \end{align*} so

Solution 2 (Change of Base)

\begin{align*} \frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x} &= 2 \\[6pt] \log_2x \cdot \log_3x &= 2(\log_2x+\log_3x) \\[6pt] \log_2x \cdot \log_3x &= 2\log_2x + 2\log_3x \\[6pt] \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} &= 2\frac{\log x}{\log 2} + 2\frac{\log x}{\log 3} \\[6pt] \frac{(\log x)^2}{\log 2 \cdot \log 3} &= \frac{2\log x \cdot \log 3 + 2\log x \cdot \log 2}{\log 2 \cdot \log 3} \\[6pt] (\log x)^2 &= 2\log x \cdot \log 3 + 2\log x \cdot \log 2 \\[6pt] (\log x)^2 &= 2\log x(\log 2 + \log 3) \\[6pt] \log x &= 2(\log 2 + \log 3) \\[6pt] x &= 10^{2(\log 2 + \log 3)} \\[6pt] x &= (10^{\log 2} \cdot 10^{\log 3})^2 \\[6pt] x &= (2 \cdot 3)^2 = 6^2 = \boxed{\textbf{(C) }36} \end{align*}

~sourodeepdeb

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.