Difference between revisions of "2005 AMC 12B Problems/Problem 13"
(→Solution) |
(→Solution) |
||
| Line 9: | Line 9: | ||
Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | ||
| + | |||
| + | ==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 12:32, 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. 
