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Difference between revisions of "1981 AHSME Problems/Problem 15"

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==Solution==
 
==Solution==
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Use rules of logarithms to solve this equation. 
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<math>(2x)^{\log_{b} 2} = (3x)^{\log_{b} 3}</math>
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<math>\log_{b} 2 \cdot \log_{b} 2x = \log_{b} 3 \cdot \log_{b} 3x</math>
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<math>\log_{b} 2 \cdot (\log_{b} 2 + \log_{b} x) = \log_{b} 3 \cdot (\log_{b} 3 + \log_{b} x)</math>
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<math>(\log_{b} 2)^2  + \log_{b} 2 \cdot \log_{b} x = (\log_{b} 3)^2 + \log_{b} 3 \cdot \log_{b} x</math>
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<math>(\log_{b} 2)^2 - (\log_{b} 3)^2  =  \log_{b} 3 \cdot \log_{b} x - \log_{b} 2 \cdot \log_{b} x </math>
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<math>\boxed {B}</math>
 
<math>\boxed {B}</math>

Revision as of 02:10, 26 June 2025

Problem

If $b>1$, $x>0$, and $(2x)^{\log_b 2}-(3x)^{\log_b 3}=0$, then $x$ is

$\textbf{(A)}\ \dfrac{1}{216}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \text{not uniquely determined}$

Solution

Use rules of logarithms to solve this equation.

$(2x)^{\log_{b} 2} = (3x)^{\log_{b} 3}$

$\log_{b} 2 \cdot \log_{b} 2x = \log_{b} 3 \cdot \log_{b} 3x$

$\log_{b} 2 \cdot (\log_{b} 2 + \log_{b} x) = \log_{b} 3 \cdot (\log_{b} 3 + \log_{b} x)$

$(\log_{b} 2)^2 + \log_{b} 2 \cdot \log_{b} x = (\log_{b} 3)^2 + \log_{b} 3 \cdot \log_{b} x$

$(\log_{b} 2)^2 - (\log_{b} 3)^2 = \log_{b} 3 \cdot \log_{b} x - \log_{b} 2 \cdot \log_{b} x$

$\boxed {B}$

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