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2024 AMC 12B Problems/Problem 8

Problem

What value of $x$ satisfies \[\frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x}=2?\]

$\textbf{(A) } 25 \qquad\textbf{(B) } 32 \qquad\textbf{(C) } 36 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 48$

Solution 1

We have log2⁡x⋅log3⁡x=2(log2⁡x+log3⁡x)1=2(log2⁡x+log3⁡x)log2⁡x⋅log3⁡x1=2(1log3⁡x+1log2⁡x)1=2(logx⁡3+logx⁡2)logx⁡6=12x12=6x=36 so $\boxed{\textbf{(C) }36}$

~kafuu_chino

Solution 2 (Change of Base)

log2⁡x⋅log3⁡xlog2⁡x+log3⁡x=2log2⁡x⋅log3⁡x=2(log2⁡x+log3⁡x)log2⁡x⋅log3⁡x=2log2⁡x+2log3⁡xlog⁡xlog⁡2⋅log⁡xlog⁡3=2log⁡xlog⁡2+2log⁡xlog⁡3(log⁡x)2log⁡2⋅log⁡3=2log⁡x⋅log⁡3+2log⁡x⋅log⁡2log⁡2⋅log⁡3(log⁡x)2=2log⁡x⋅log⁡3+2log⁡x⋅log⁡2(log⁡x)2=2log⁡x(log⁡2+log⁡3)log⁡x=2(log⁡2+log⁡3)x=102(log⁡2+log⁡3)x=(10log⁡2⋅10log⁡3)2x=(2⋅3)2=62=(C) 36

~sourodeepdeb

Solution 3 (Using Variables)

Let $a=\log_2x$ and $b=\log_3x$. This gives us the equation \[\frac{ab}{a+b}=2.\]

Then, from our definitions of $a$ and $b$, $2^a=x$ and $3^b=x$, so $2^a=3^b.$ Taking the logarithm base $3$ of both sides of this equation gives us $\log_3 2^a=b$, hence $a \log_3 2=b.$ Now, we substitute $a \log_3 2$ for $b$ in the equation, which gives \[\frac{a \cdot a \log_3 2}{a+a \log_3 2}=2.\]Notice that we can factor out an $a$ in the numerator and denominator, if $a \neq 0,$ and doing so yields \[\frac{a \log_3 2}{1+\log_3 2}=2.\] We know that $1= \log_3 3,$ so putting that in gives us \[\frac{a \log_3 2}{\log_3 3+\log_3 2}=2 \implies \frac{a \log_3 2}{\log_3 6}=2.\]So, $a=2 \cdot \frac{\log_3 6}{\log_3 2}$, which, using the change of base formula, is equivalent to $2 \cdot \log_2 6,$ thus, \[a=2 \cdot \log_2 6= \log _2 6^2= \log _2 36.\] Finally, using our original definition of $a,$ we have \[a = \log_2 x=\log_2 36,\] so $x=\boxed{\textbf{(C) }36}.$

~hdanger

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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