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AIME 2010 I · 第 3 题

AIME 2010 I — Problem 3

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Suppose that y=34xy = \frac34x and xy=yxx^y = y^x. The quantity x+yx + y can be expressed as a rational number rs\frac {r}{s}, where rr and ss are relatively prime positive integers. Find r+sr + s.

解析

Solution 1

Substitute y=34xy = \frac34x into xy=yxx^y = y^x and solve.

x34x=(34x)xx^{\frac34x} = \left(\frac34x\right)^x x34x=(34)xxxx^{\frac34x} = \left(\frac34\right)^x \cdot x^x x14x=(34)xx^{-\frac14x} = \left(\frac34\right)^x x14=34x^{-\frac14} = \frac34 x=25681x = \frac{256}{81} y=34x=19281y = \frac34x = \frac{192}{81} x+y=44881x + y = \frac{448}{81} 448+81=529448 + 81 = \boxed{529}

Solution 2

We solve in general using cc instead of 3/43/4. Substituting y=cxy = cx, we have:

xcx=(cx)x(xx)c=cxxxx^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x Dividing by xxx^x, we get (xx)c1=cx(x^x)^{c - 1} = c^x.

Taking the xxth root, xc1=cx^{c - 1} = c, or x=c1/(c1)x = c^{1/(c - 1)}.

In the case c=34c = \frac34, x=(34)4=(43)4=25681x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}, y=6427y = \frac {64}{27}, x+y=256+19281=44881x + y = \frac {256 + 192}{81} = \frac {448}{81}, yielding an answer of 448+81=529448 + 81 = \boxed{529}.

Solution 3

Taking the logarithm base xx of both sides, we arrive with:

y=logxyxyx=logxy=logx34x=34y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \frac{3}{4}x = \frac{3}{4} Where the last two simplifications were made since y=34xy = \frac{3}{4}x. Then,

x34=34xx14=43x=(43)4x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4 Then, y=(43)3y = \left(\frac{4}{3}\right)^3, and thus:

x+y=(43)3(43+1)=44881448+81=529x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}

Solution 4 (another version of Solution 3)

Taking the logarithm base xx of both sides, we arrive with:

y=logxyxyx=logxy=logx(34x)=34y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4} Now we proceed by the logarithm rule log(ab)=loga+logb\log(ab)=\log a + \log b. The equation becomes:

logx34+logxx=34\log_x \frac{3}{4} + \log_x x = \frac{3}{4} logx34+1=34\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4} logx34=14\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4} x14=34\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4} 1x14=34\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4} x14=43\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3} x4=43\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3} x=(43)4=25681\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81} Then find yy as in solution 3, and we get 529\boxed{529}.