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Difference between revisions of "2002 AMC 12P Problems/Problem 21"

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== Solution ==
 
== Solution ==
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We may rewrite the given equation as <cmath>2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}</cmath>
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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>
 
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>
  
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Factoring gives <math>(2x-y)(x-2y) = 0</math> or <math>\frac {x}{y} = 2, \frac {1}{2}</math>.
 
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
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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

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