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Difference between revisions of "2024 AMC 12B Problems/Problem 8"

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We have
 
We have
 
\begin{align*}
 
\begin{align*}
−
&\log_2x\cdot\log_3x=2(\log_2x+\log_3x) \\
+
\log_2x\cdot\log_3x&=2(\log_2x+\log_3x) \\
−
&1=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\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(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\
−
&1=2(\log_x3+\log_x2) \\
+
1&=2(\log_x3+\log_x2) \\
−
&\log_x6=\frac{1}{2} \\
+
\log_x6&=\frac{1}{2} \\
−
&x^{\frac{1}{2}}=6 \\
+
x^{\frac{1}{2}}&=6 \\
−
&x=36
+
x&=36
 
\end{align*}
 
\end{align*}
 
so <math>\boxed{\textbf{(C) }36}</math>
 
so <math>\boxed{\textbf{(C) }36}</math>
 +
 +
~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 <math>a=\log_2x</math> and <math>b=\log_3x</math>. This gives us the equation <cmath>\frac{ab}{a+b}=2.</cmath>
 +
 +
Then, from our definitions of <math>a</math> and <math>b</math>, <math>2^a=x</math> and <math>3^b=x</math>, so <math>2^a=3^b.</math> Taking the logarithm base <math>3</math> of both sides of this equation gives us <math>\log_3 2^a=b</math>, hence <math>a \log_3 2=b.</math>
 +
Now, we substitute <math>a \log_3 2</math> for <math>b</math> in the equation, which gives <cmath>\frac{a \cdot a \log_3 2}{a+a \log_3 2}=2.</cmath>Notice that we can factor out an <math>a</math> in the numerator and denominator, if <math>a \neq 0,</math> and doing so yields <cmath>\frac{a \log_3 2}{1+\log_3 2}=2.</cmath> We know that <math>1= \log_3 3,</math> so putting that in gives us <cmath>\frac{a \log_3 2}{\log_3 3+\log_3 2}=2 \implies \frac{a \log_3 2}{\log_3 6}=2.</cmath>So, <math>a=2 \cdot  \frac{\log_3 6}{\log_3 2}</math>, which, using the change of base formula, is equivalent to <math>2 \cdot \log_2 6,</math> thus, <cmath>a=2 \cdot \log_2 6= \log _2 6^2= \log _2 36.</cmath> Finally, using our original definition of <math>a,</math> we have <cmath>a = \log_2 x=\log_2 36,</cmath> so <math>x=\boxed{\textbf{(C) }36}.</math>
 +
 +
~hdanger
 +
 +
==Video Solution 1 by TheBeautyofMath==
 +
https://youtu.be/AKLPjTRPF4Q?t=539
 +
 +
~IceMatrix
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2024|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:28, 6 October 2025

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, 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

Video Solution 1 by TheBeautyofMath

https://youtu.be/AKLPjTRPF4Q?t=539

~IceMatrix

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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