1957 AHSME Problems/Problem 5

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 $\boxed{\textbf{(B)} \log\frac xy}.$

Page

Toolbox

Search

1957 AHSME Problems/Problem 5

Solution