Difference between revisions of "1950 AHSME Problems/Problem 25"
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<math> \log_{5}\frac{(125)(625)}{25} </math> can be simplified to <math> \log_{5}\ (125)(25) </math> since <math>25^2 = 625</math>. <math>125 = 5^3</math> and <math>5^2 = 25</math> so <math> \log_{5}\ 5^5 </math> would be the simplest form. In <math> \log_{5}\ 5^5 </math>, <math>5^x = 5^5</math>. Therefore, <math>x = 5</math> and the answer is <math>\boxed{\mathrm{(D)}\ 5}</math> | <math> \log_{5}\frac{(125)(625)}{25} </math> can be simplified to <math> \log_{5}\ (125)(25) </math> since <math>25^2 = 625</math>. <math>125 = 5^3</math> and <math>5^2 = 25</math> so <math> \log_{5}\ 5^5 </math> would be the simplest form. In <math> \log_{5}\ 5^5 </math>, <math>5^x = 5^5</math>. Therefore, <math>x = 5</math> and the answer is <math>\boxed{\mathrm{(D)}\ 5}</math> | ||
| − | {{AHSME box|year=1950|num-b=24|num-a=26}} | + | == See Also == |
| + | {{AHSME 50p box|year=1950|num-b=24|num-a=26}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 07:34, 29 April 2012
Problem
The value of 
is equal to:
Solution
can be simplified to 
since 
. 
and 
so 
would be the simplest form. In , 
. Therefore, 
and the answer is 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
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| All AHSME Problems and Solutions | ||
