Difference between revisions of "2005 AMC 12B Problems/Problem 13"
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Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | ||
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+ | ==Alternate Solution== | ||
+ | Changing <math>4^{x_1}=5</math> to logarithmic form, we get <math>{x_1}=log_4 5</math>. We can rewrite this as <math>{x_1}=\dfrac{log_5}{log_4}</math>. Applying this to the rest, we get <math>x_1x_2...x_{124}=\dfrac{log5}{log4}*\dfrac{log6}{log5}*...*\dfrac{log128}{log127}=\dfrac{log5log6...log128}{log4log5...log127}=\dfrac{log128}{log4}=log_4128=log_4{2^7}=7*log_42=7*\dfrac{1}{2}=\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2005|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2005|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:32, 31 December 2016
Problem
Suppose that , , , ... , . What is ?
Solution
We see that we can re-write , , , ... , as by using substitution. By using the properties of exponents, we know that .
Therefore, the answer is
Alternate Solution
Changing to logarithmic form, we get . We can rewrite this as . Applying this to the rest, we get
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.