Difference between revisions of "1950 AHSME Problems/Problem 25"
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<math> \textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these} </math> | <math> \textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these} </math> | ||
− | == Solution == | + | == Solution 1== |
<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> | ||
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+ | ==Solution 2== | ||
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+ | <math> \log_{5}\frac{(125)(625)}{25} </math> can be also represented as <math> \log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5 </math> which can be solved to get <math>5</math> | ||
== See Also == | == See Also == |
Revision as of 13:05, 15 January 2018
Contents
[hide]Problem
The value of is equal to:
Solution 1
can be simplified to since . and so would be the simplest form. In , . Therefore, and the answer is
Solution 2
can be also represented as which can be solved to get
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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