Difference between revisions of "1959 AHSME Problems/Problem 45"

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== Solution ==
 
== Solution ==
<math>\box{\textbf{(B)}}</math>
+
From the properties of [[logarithms]], we can simplify the equation and solve for <math>y</math>:
 +
\begin{align*}
 +
(\log_3 x)(\log_x 2x)(\log_{2x} y) &= \log_{x}x^2 \\
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(\log_3 2x)(\log_{2x} y) &= 2\log_x x \\
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\log_3 y &= 2 \\
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y &= 3^2 \\
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y &= 9
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\end{align*}
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Thus, our answer is <math>\boxed{\textbf{(B) }9}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=44|num-a=46}}
 
{{AHSME 50p box|year=1959|num-b=44|num-a=46}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:41, 22 July 2024

Problem

If $\left(\log_3 x\right)\left(\log_x 2x\right)\left( \log_{2x} y\right)=\log_{x}x^2$, then $y$ equals:

$\textbf{(A)}\ \frac92\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 81$

Solution

From the properties of logarithms, we can simplify the equation and solve for $y$: \begin{align*} (\log_3 x)(\log_x 2x)(\log_{2x} y) &= \log_{x}x^2 \\ (\log_3 2x)(\log_{2x} y) &= 2\log_x x \\ \log_3 y &= 2 \\ y &= 3^2 \\ y &= 9 \end{align*} Thus, our answer is $\boxed{\textbf{(B) }9}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 44
Followed by
Problem 46
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All AHSME Problems and Solutions

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