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2024 AMC 12B Problems/Problem 8

Revision as of 12:42, 18 November 2024 by Hdanger (talk | contribs) (→‎Solution 3 (Using Variables))

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 \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 $\boxed{\textbf{(C) }36}$

~kafuu_chino

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

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 in reverse, is equivalent to $2 \cdot \log_2 6,$ thus, \[a= \log _2 6^2= \log _2 36.\] 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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