Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1957 AHSME Problems/Problem 5"
m (see also box)
 
Line 20: Line 20:
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
so the answer is <math>\boxed{\textbf{(B)} \log\frac xy}.</math>
 
so the answer is <math>\boxed{\textbf{(B)} \log\frac xy}.</math>
 +
 +
== See also ==
 +
 +
{{AHSME 50p box|year=1957|num-b=4|num-a=6}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:04, 25 July 2024

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

See also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_5&oldid=223577"