2009 AMC 12B Problems/Problem 15
Problem
Assume . Below are five equations for . Which equation has the largest solution ?
Solution
(B) Intuitively, will be largest for that option for which the value in the parentheses is smallest.
Formally, first note that each of the values in parentheses is larger than . Now, each of the options is of the form . This can be rewritten as . As , we have . Thus is the largest for the option for which is smallest. And as is an increasing function, this is the option for which is smallest.
We now get the following easier problem: Given that , find the smallest value in the set .
Clearly is smaller than the first and the third option.
We have , dividing both sides by we get .
And finally, , therefore , and as both sides are positive, we can take the square root and get .
Thus the answer is .
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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.