1957 AHSME Problems/Problem 5

Problem 5

Through the use of theorems on logarithms $\log{\frac{a}{b}} + \log{\frac{b}{c}} + \log{\frac{c}{d}} - \log{\frac{ay}{dx}}$ can be reduced to:

$\textbf{(A)}\ \log{\frac{y}{x}}\qquad \textbf{(B)}\ \log{\frac{x}{y}}\qquad \textbf{(C)}\ 1\qquad \\ \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}$

Solution

Using the properties $\log(x)+\log(y)=\log(xy)$ and $\log(x)-\log(y)=\log(x/y)$ , we have $\begin{align*} \log\frac ab+\log\frac bc+\log\frac cd-\log\frac{ay}{dx}&=\log\left(\frac ab\cdot\frac bc\cdot\frac cd\right)-\log \frac {ay}{dx} \\ &=\log \frac ad-\log\frac{ay}{dx} \\ &=\log\left(\frac{\frac ad}{\frac{ay}{dx}}\right) \\ &=\log \frac xy, \end{align*}$ so the answer is $\boxed{\textbf{(B)} \log\frac xy}.$

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1957 AHSME Problems/Problem 5

Problem 5

Solution

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