Difference between revisions of "2010 AIME I Problems/Problem 3"
m (Semi-automated contest formatting - script by azjps) |
Mathfun1000 (talk | contribs) m |
||
(15 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | Suppose that <math>y = \frac34x</math> and <math>x^y = y^x</math>. The quantity <math>x + y</math> can be expressed as a rational number <math>\frac {r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r + s</math>. | ||
− | == See | + | == Solution 1 == |
+ | Substitute <math>y = \frac34x</math> into <math>x^y = y^x</math> and solve. | ||
+ | <cmath>x^{\frac34x} = \left(\frac34x\right)^x</cmath> | ||
+ | <cmath>x^{\frac34x} = \left(\frac34\right)^x \cdot x^x</cmath> | ||
+ | <cmath>x^{-\frac14x} = \left(\frac34\right)^x</cmath> | ||
+ | <cmath>x^{-\frac14} = \frac34</cmath> | ||
+ | <cmath>x = \frac{256}{81}</cmath> | ||
+ | <cmath>y = \frac34x = \frac{192}{81}</cmath> | ||
+ | <cmath>x + y = \frac{448}{81}</cmath> | ||
+ | <cmath>448 + 81 = \boxed{529}</cmath> | ||
+ | |||
+ | == Solution 2 == | ||
+ | We solve in general using <math>c</math> instead of <math>3/4</math>. Substituting <math>y = cx</math>, we have: | ||
+ | |||
+ | <center><cmath>x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x</cmath></center> | ||
+ | |||
+ | Dividing by <math>x^x</math>, we get <math>(x^x)^{c - 1} = c^x</math>. | ||
+ | |||
+ | Taking the <math>x</math>th root, <math>x^{c - 1} = c</math>, or <math>x = c^{1/(c - 1)}</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 3 == | ||
+ | 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 = \left(\frac{4}{3}\right)^4</cmath></center> | ||
+ | Then, <math>y = \left(\frac{4}{3}\right)^3</math>, and thus: | ||
+ | <center> <cmath>x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}</cmath> </center> | ||
+ | |||
+ | == Solution 4 (another version of Solution 3)== | ||
+ | Taking the logarithm base <math>x</math> of both sides, we arrive with: | ||
+ | <cmath> y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}</cmath> | ||
+ | Now we proceed by the logarithm rule <math>\log(ab)=\log a + \log b</math>. The equation becomes: | ||
+ | <cmath>\log_x \frac{3}{4} + \log_x x = \frac{3}{4}</cmath> | ||
+ | <cmath>\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}</cmath> | ||
+ | <cmath>\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}</cmath> | ||
+ | <cmath>\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}</cmath> | ||
+ | <cmath>\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}</cmath> | ||
+ | <cmath>\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}</cmath> | ||
+ | <cmath>\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}</cmath> | ||
+ | <cmath>\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}</cmath> | ||
+ | Then find <math>y</math> as in solution 3, and we get <math>\boxed{529}</math>. | ||
+ | |||
+ | == See Also == | ||
{{AIME box|year=2010|num-b=2|num-a=4|n=I}} | {{AIME box|year=2010|num-b=2|num-a=4|n=I}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:07, 17 December 2021
Contents
Problem
Suppose that and . The quantity can be expressed as a rational number , where and are relatively prime positive integers. Find .
Solution 1
Substitute into and solve.
Solution 2
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 3
Taking the logarithm base of both sides, we arrive with:
Where the last two simplifications were made since . Then,
Then, , and thus:
Solution 4 (another version of Solution 3)
Taking the logarithm base of both sides, we arrive with: Now we proceed by the logarithm rule . The equation becomes: Then find as in solution 3, and we get .
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.