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

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 $\boxed {\text{(C) }2}$ .

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 1: / Line 1: @@
 == Problem ==
-How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+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
-<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }  </math>
+<cmath>2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.</cmath>
+Find the largest possible value of <math>\log_a b.</math>
+<math>
+\text{(A) }\sqrt{2}
+\qquad
+\text{(B) }\sqrt{3}
+\qquad
+\text{(C) }2
+\qquad
+\text{(D) }\sqrt{6}
+\qquad
+\text{(E) }3
+</math>
 == 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 ==
-{{AMC12 box|year=2000|num-b=6|num-a=8}}
+{{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"

Latest revision as of 21:26, 26 July 2024

Problem

Solution

See also