Simple and elegant

Algebra Level 3

Out of the following options, which is a solution to:

x y = y x x^y=y^x and x 2 = y 3 x^2=y^3

2,3 3,2 27/4,9/8 27/8, 9/4

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1 solution

Tom Engelsman
Feb 9, 2017

Taking x 2 = y 3 y = x 2 3 , x^2 = y^3 \Rightarrow y = x^{\frac{2}{3}}, substituting this value into other equation will yield:

x y = y x x x 2 3 = ( x 2 3 ) x ; x^y = y^x \Rightarrow x^{x^{\frac{2}{3}}} = (x^{\frac{2}{3}})^x;

or x 2 3 = 2 x 3 x 2 = 8 x 3 27 ; x^{\frac{2}{3}} = \frac{2x}{3} \Rightarrow x^2 = \frac{8x^3}{27};

or 8 x 3 27 x 2 = 0 x 2 ( 8 x 27 ) = 0 x = 0 , 27 8 8x^3 - 27x^2 = 0 \Rightarrow x^{2}(8x - 27) = 0 \Rightarrow x = 0, \frac{27}{8} and y = 0 , ( 27 8 ) 2 3 = 9 4 . y = 0, (\frac{27}{8})^{\frac{2}{3}} = \frac{9}{4}.

Hence the solutions are ( x , y ) = ( 0 , 0 ) ; ( 27 8 , 9 4 ) . (x,y) = (0, 0); (\frac{27}{8}, \frac{9}{4}). But 0 0 0^{0} is indeterminate, which only leaves x = 27 8 , y = 9 4 . \boxed{x = \frac{27}{8}, y = \frac{9}{4}}.

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