Difference between revisions of "2010 AIME I Problems/Problem 3"
(→Solution) |
|||
Line 12: | Line 12: | ||
In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>. | In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Taking the logarithm base <math>x</math> of both sides, we arrive with: | ||
+ | |||
+ | <center><cmath> y = log_x y^x \Longrightarrow \frac{y}{x} = log_x y = log_x \frac{3}{4}x = \frac{3}{4}</cmath></center> | ||
+ | Where the last two simplifications were made since <math>y = \frac{3}{4}x</math>. Then, | ||
+ | <center><cmath>x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = (\frac{4}{3})^4</cmath></center> | ||
+ | Then, <math>y = (\frac{4}{3})^3</math>, and thus: | ||
+ | <center> <cmath>x+y = (\frac{4}{3})^3 (\frac{4}{3} + 1) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}</cmath> </center> | ||
== See Also == | == See Also == |
Revision as of 19:25, 28 February 2016
Contents
[hide]Problem
Suppose that and . The quantity can be expressed as a rational number , where and are relatively prime positive integers. Find .
Solution
We solve in general using instead of . Substituting , we have:
Dividing by , we get .
Taking the th root, , or .
In the case , , , , yielding an answer of .
Solution 2
Taking the logarithm base of both sides, we arrive with:
Where the last two simplifications were made since . Then,
Then, , and thus:
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.