Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1950 AHSME Problems/Problem 25"
m (small latex fix)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
==Problem==
+
== Problem ==
 
The value of <math> \log_{5}\frac{(125)(625)}{25} </math> is equal to:
 
The value of <math> \log_{5}\frac{(125)(625)}{25} </math> is equal to:
  
 
<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 ==
+
== 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>.
  
−
<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 ==
 
== See Also ==
 
{{AHSME 50p box|year=1950|num-b=24|num-a=26}}
 
{{AHSME 50p box|year=1950|num-b=24|num-a=26}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:58, 11 October 2020

Problem

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

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

Solutions

Solution 1

$\log_{5}\frac{(125)(625)}{25}$ 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 $\boxed{\mathrm{(D)}\ 5}$.

Solution 2

$\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{\mathrm{(D)}\ 5}$.

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_25&oldid=134944"