Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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==Solution using chain logarithm rule== | ==Solution using chain logarithm rule== | ||
− | As in solution 2, 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>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer. | + | As in solution 2, 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>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer. |
+ | ~solver1104 | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:55, 6 January 2019
Contents
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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