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

m
Line 9: Line 9:
  
 
{{AHSME box|year=1950|num-b=24|num-a=26}}
 
{{AHSME box|year=1950|num-b=24|num-a=26}}
 +
[[Category:Introductory Algebra Problems]]

Revision as of 14:30, 17 April 2012

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 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions