Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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We can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_5 6</math>, <math>c</math> as <math>\log_6 7</math>, and <math>d</math> as <math>\log_7 8</math>. | We can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_5 6</math>, <math>c</math> as <math>\log_6 7</math>, and <math>d</math> as <math>\log_7 8</math>. | ||
We know that <math>\log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}</math>, so <math>a*b*c*d=</math> | We know that <math>\log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}</math>, so <math>a*b*c*d=</math> | ||
− | <cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}</cmath> | + | <cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}=</cmath> |
<cmath>\frac{\log8}{\log4}=</cmath> | <cmath>\frac{\log8}{\log4}=</cmath> | ||
− | <cmath>\frac{3\log2}{2\log2}</cmath> | + | <cmath>\frac{3\log2}{2\log2}=</cmath> |
<cmath>\boxed{\frac{3}{2}}</cmath> | <cmath>\boxed{\frac{3}{2}}</cmath> |
Revision as of 21:58, 31 May 2021
Contents
[hide]Problem
Suppose that , , , and . What is ?
Solution
Solution using logarithms
We can write as , as , as , and as . We know that can be rewritten as , so
Solution using chain logarithm rule
As in solution 2, we can write as , as , as , and as . is equivalent to . Note that by the logarithm chain rule, this is equivalent to , which evaluates to , so is the answer. ~solver1104
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
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