Difference between revisions of "1957 AHSME Problems/Problem 5"
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+ | == Problem 5== | ||
+ | |||
+ | Through the use of theorems on logarithms | ||
+ | <cmath>\log{\frac{a}{b}} + \log{\frac{b}{c}} + \log{\frac{c}{d}} - \log{\frac{ay}{dx}} </cmath> | ||
+ | can be reduced to: | ||
+ | |||
+ | <math>\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} </math> | ||
+ | |||
==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 | ||
Line 8: | 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 can be reduced to:
Solution
Using the properties and , we have so the answer is
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 |
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