Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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<cmath> 8=7^d </cmath> <cmath>8=\left(6^c\right)^d</cmath> <cmath>8=\left(\left(5^b\right)^c\right)^d</cmath> <cmath>8=\left(\left(\left(4^a\right)^b\right)^c\right)^d</cmath> <cmath>8=4^{a\cdot b\cdot c\cdot d}</cmath> <cmath>2^3=2^{2\cdot a\cdot b\cdot c\cdot d}</cmath> <cmath>3=2\cdot a\cdot b\cdot c\cdot d</cmath> <cmath>a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}</cmath> | <cmath> 8=7^d </cmath> <cmath>8=\left(6^c\right)^d</cmath> <cmath>8=\left(\left(5^b\right)^c\right)^d</cmath> <cmath>8=\left(\left(\left(4^a\right)^b\right)^c\right)^d</cmath> <cmath>8=4^{a\cdot b\cdot c\cdot d}</cmath> <cmath>2^3=2^{2\cdot a\cdot b\cdot c\cdot d}</cmath> <cmath>3=2\cdot a\cdot b\cdot c\cdot d</cmath> <cmath>a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}</cmath> | ||
+ | ==Solution using [[logarithms]]== | ||
+ | We can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</math>. | ||
+ | We know that <math>log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}, so </math>a*b*c*d=$ | ||
+ | <cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}</cmath> | ||
+ | |||
+ | <cmath>\frac{\log8}{\log4}</cmath> | ||
+ | |||
+ | <cmath>\frac{3\log2}{2\log2}</cmath> | ||
+ | |||
+ | <cmath>\boxed{\frac{3}{2}}</cmath> | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}} |
Revision as of 22:30, 24 May 2013
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 a*b*c*d=$
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |