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 "1970 AHSME Problems/Problem 8"

m (→‎See also)
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
If <math>a=log_8(225)</math> and <math>b=log_2(15)</math>, then
+
If <math>a=\log_8 225</math> and <math>b=\log_2 15</math>, then
  
 
<math>\text{(A) } a=b/2\quad
 
<math>\text{(A) } a=b/2\quad
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
The solutions imply that finding the ratio <math>\frac{a}{b}</math> will solve the problem.  We compute <math>\frac{a}{b}</math>, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:
 +
 
 +
<math>\frac{\log_8 225}{\log_2 15}</math>
 +
 
 +
<math>\frac{\frac{\ln 225}{\ln 8}}{\frac{\ln 15}{\ln 2}}</math>
 +
 
 +
<math>\frac{\ln 225 \ln 2}{\ln 15 \ln 8}</math>
 +
 
 +
<math>\frac{\ln 225}{\ln 15} \cdot \frac{\ln 2}{\ln 8}</math>
 +
 
 +
<math>\ln_{15} 225 \cdot \ln_8 2</math>
 +
 
 +
Since <math>15^2 = 225</math>, the first logarithm is <math>2</math>.  Since <math>8^{\frac{1}{3}} = 2</math>, the second logarithm is <math>\frac{1}{3}</math>.
 +
 
 +
Thus, we have <math>\frac{a}{b} = \frac{2}{3}</math>, or <math>a = \frac{2}{3}b</math>, which is option <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 20:02, 13 July 2019

Problem

If $a=\log_8 225$ and $b=\log_2 15$, then

$\text{(A) } a=b/2\quad \text{(B) } a=2b/3\quad \text{(C) } a=b\quad \text{(D) } b=a/2\quad \text{(E) } a=3b/2$

Solution

The solutions imply that finding the ratio $\frac{a}{b}$ will solve the problem. We compute $\frac{a}{b}$, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:

$\frac{\log_8 225}{\log_2 15}$

$\frac{\frac{\ln 225}{\ln 8}}{\frac{\ln 15}{\ln 2}}$

$\frac{\ln 225 \ln 2}{\ln 15 \ln 8}$

$\frac{\ln 225}{\ln 15} \cdot \frac{\ln 2}{\ln 8}$

$\ln_{15} 225 \cdot \ln_8 2$

Since $15^2 = 225$, the first logarithm is $2$. Since $8^{\frac{1}{3}} = 2$, the second logarithm is $\frac{1}{3}$.

Thus, we have $\frac{a}{b} = \frac{2}{3}$, or $a = \frac{2}{3}b$, which is option $\fbox{B}$.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
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=1970_AHSME_Problems/Problem_8&oldid=107606"