Difference between revisions of "2005 AMC 12B Problems/Problem 13"
(→Solution) |
(→Solution) |
||
| Line 6: | Line 6: | ||
We see that we can re-write <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math> as <math>\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128</math> by using substitution. By using the properties of exponents, we know that <math>4^{x_1x_2...x_{124}}=128</math>. | We see that we can re-write <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math> as <math>\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128</math> by using substitution. By using the properties of exponents, we know that <math>4^{x_1x_2...x_{124}}=128</math>. | ||
| − | <math>4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2 | + | <math>4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}</math> |
Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | ||
Revision as of 12:35, 20 June 2015
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 
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. 
