1959 AHSME Problems/Problem 45

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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