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Difference between revisions of "1957 AHSME Problems/Problem 5"

(Created page with "Using the properties <math>\log(x)+\log(y)=\log(xy)</math> and <math>\log(x)-\log(y)=\log(x/y)</math>, we have <cmath>\begin{align*} \log\frac ab+\log\frac bc+\log\frac cd-\lo...")
 
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==Solution==
 
Using the properties <math>\log(x)+\log(y)=\log(xy)</math> and <math>\log(x)-\log(y)=\log(x/y)</math>, we have
 
Using the properties <math>\log(x)+\log(y)=\log(xy)</math> and <math>\log(x)-\log(y)=\log(x/y)</math>, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}

Revision as of 23:19, 3 January 2019

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

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