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

Revision as of 14:58, 10 March 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 $2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}$ 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 $2x^2 - 5xy + 2y^2 = 0$

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.

@@ Line 23: / Line 23: @@
 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>
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=20|num-a=22}}
 {{MAA Notice}}

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

Revision as of 14:58, 10 March 2024

Problem

Solution

See also