Difference between revisions of "1954 AHSME Problems/Problem 15"

(Created page with "== Problem 15== <math>\log 125</math> equals: <math>\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \ \textbf{(D)}\ 3 - 3\log 2 \...")
 
 
Line 8: Line 8:
 
== Solution ==
 
== Solution ==
 
<math>\log(1000)=\log(125)+\log(8)\implies 3=\log(125)+\log(2^3)\implies \log(125)=3-3\log(2)</math>, <math>\fbox{D}</math>
 
<math>\log(1000)=\log(125)+\log(8)\implies 3=\log(125)+\log(2^3)\implies \log(125)=3-3\log(2)</math>, <math>\fbox{D}</math>
 +
 +
==See Also==
 +
 +
{{AHSME 50p box|year=1954|num-b=14|num-a=16}}
 +
 +
{{MAA Notice}}

Latest revision as of 00:26, 28 February 2020

Problem 15

$\log 125$ equals:

$\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25  \\ \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$

Solution

$\log(1000)=\log(125)+\log(8)\implies 3=\log(125)+\log(2^3)\implies \log(125)=3-3\log(2)$, $\fbox{D}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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