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 "2002 AMC 12P Problems/Problem 21"

(→‎Problem)
m (→‎Solution)
 
(16 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
Let <math>a</math> and <math>b</math> be real numbers greater than <math>1</math> for which there exists a positive real number <math>c,</math> different from <math>1</math>, such that
 
Let <math>a</math> and <math>b</math> be real numbers greater than <math>1</math> for which there exists a positive real number <math>c,</math> different from <math>1</math>, such that
 +
 
<cmath>2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.</cmath>
 
<cmath>2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.</cmath>
 +
 
Find the largest possible value of <math>\log_a b.</math>
 
Find the largest possible value of <math>\log_a b.</math>
  
Line 17: Line 19:
  
 
== Solution ==
 
== Solution ==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
We may rewrite the given equation as <cmath>2 \left(\frac {\log c}{\log a} + \frac {\log c}{\log b} \right) = \frac {9\log c}{\log a + \log b}</cmath>
 +
Since <math>c \neq 1</math>, we have <math>\log c \neq 0</math>, so we may divide by <math>\log c</math> on both sides. After making the substitutions <math>x = \log a</math> and <math>y = \log b</math>, our equation becomes <cmath>\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}</cmath>
 +
 
 +
Rewriting the left-hand side gives <cmath>\frac {2(x+y)}{xy} = \frac {9}{x+y}</cmath>
 +
 
 +
Cross-multiplying gives <math>2(x+y)^2 = 9xy</math> or <cmath>2x^2 - 5xy + 2y^2 = 0</cmath>
 +
 
 +
Factoring gives <math>(2x-y)(x-2y) = 0</math> or <math>\frac {x}{y} = 2, \frac {1}{2}</math>.
 +
 
 +
Recall that <math>\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b</math>. Therefore, the maximum value of <math>\log_{a} b</math> is <math>\boxed {\text{(C) }2}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=20|num-a=22}}
 
{{AMC12 box|year=2002|ab=P|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:26, 26 July 2024

Problem

Let $a$ and $b$ be real numbers greater than $1$ for which there exists a positive real number $c,$ different from $1$, such that

\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]

Find the largest possible value of $\log_a b.$

$\text{(A) }\sqrt{2} \qquad \text{(B) }\sqrt{3} \qquad \text{(C) }2 \qquad \text{(D) }\sqrt{6} \qquad \text{(E) }3$

Solution

We may rewrite the given equation as \[2 \left(\frac {\log c}{\log a} + \frac {\log c}{\log b} \right) = \frac {9\log c}{\log a + \log b}\] Since $c \neq 1$, we have $\log c \neq 0$, so we may divide by $\log c$ on both sides. After making the substitutions $x = \log a$ and $y = \log b$, our equation becomes \[\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}\]

Rewriting the left-hand side gives \[\frac {2(x+y)}{xy} = \frac {9}{x+y}\]

Cross-multiplying gives $2(x+y)^2 = 9xy$ or \[2x^2 - 5xy + 2y^2 = 0\]

Factoring gives $(2x-y)(x-2y) = 0$ or $\frac {x}{y} = 2, \frac {1}{2}$.

Recall that $\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b$. Therefore, the maximum value of $\log_{a} b$ is $\boxed {\text{(C) }2}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 12 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=2002_AMC_12P_Problems/Problem_21&oldid=223731"