1950 AHSME Problems/Problem 25: Difference between revisions
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== | == Problem == | ||
The value of <math> \log_{5}\frac{(125)(625)}{25} </math> is equal to: | |||
<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> \textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these} </math> | ||
== Solutions == | |||
=== 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>. | |||
=== Solution 2 === | |||
<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>\boxed{\mathrm{(D)}\ 5}</math>. | |||
== See Also == | |||
{{AHSME 50p box|year=1950|num-b=24|num-a=26}} | |||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 23:58, 11 October 2020
Problem
The value of
is equal to:
Solutions
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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| All AHSME Problems and Solutions | ||
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