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

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== Problem 5==
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Through the use of theorems on logarithms
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<cmath>\log{\frac{a}{b}} + \log{\frac{b}{c}} + \log{\frac{c}{d}} - \log{\frac{ay}{dx}} </cmath>
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can be reduced to:
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<math>\textbf{(A)}\ \log{\frac{y}{x}}\qquad
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\textbf{(B)}\ \log{\frac{x}{y}}\qquad
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\textbf{(C)}\ 1\qquad \\
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\textbf{(D)}\ 140x-24x^2+x^3\qquad
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\textbf{(E)}\ \text{none of these}  </math>
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==Solution==
 
==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

Revision as of 23:20, 3 January 2019

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

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