1981 AHSME Problems/Problem 15: Difference between revisions

Revision as of 01:10, 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.

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

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

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

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

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

$\boxed {B}$

@@ Line 6: / Line 6: @@
 ==Solution==
+Use rules of logarithms to solve this equation.
+<math>(2x)^{\log_{b} 2} = (3x)^{\log_{b} 3}</math>
+<math>\log_{b} 2 \cdot \log_{b} 2x = \log_{b} 3 \cdot \log_{b} 3x</math>
+<math>\log_{b} 2 \cdot (\log_{b} 2 + \log_{b} x) = \log_{b} 3 \cdot (\log_{b} 3 + \log_{b} x)</math>
+<math>(\log_{b} 2)^2  + \log_{b} 2 \cdot \log_{b} x = (\log_{b} 3)^2 + \log_{b} 3 \cdot \log_{b} x</math>
+<math>(\log_{b} 2)^2 - (\log_{b} 3)^2  =  \log_{b} 3 \cdot \log_{b} x - \log_{b} 2 \cdot \log_{b} x </math>
 <math>\boxed {B}</math>

Page

Toolbox

Search

1981 AHSME Problems/Problem 15: Difference between revisions

Revision as of 01:10, 26 June 2025

Problem

Solution