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

m
m (small latex fix)
 
(3 intermediate revisions by 3 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>.
  
{{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]]
 +
{{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 (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 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