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

Latest revision as of 14:03, 28 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.

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

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

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

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

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

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

$\log_{b} 2 + \log_{b} 3 = -\log_{b} x$

$\log_{b} 6 = -\log_{b} x$

$6 = \frac{1}{x}$

$x = \frac{1}{6}\ \fbox {(B)}$

-j314andrews

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 -j314andrews
+==See also==
+{{AHSME box|year=1981|num-b=14|num-a=16}}
+{{MAA Notice}}

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

Latest revision as of 14:03, 28 June 2025

Problem

Solution

See also