Difference between revisions of "1950 AHSME Problems/Problem 25"

Revision as of 10:59, 5 July 2013

Problem

The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:

$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$

Solution

$\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ . $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ , $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{\mathrm{(D)}\ 5}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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.

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 {{AHSME 50p box|year=1950|num-b=24|num-a=26}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1950 AHSME Problems/Problem 25"

Revision as of 10:59, 5 July 2013

Problem

Solution

See Also