Solution 1
Substitute y = 3 4 x y = \frac34x y = 4 3 x into x y = y x x^y = y^x x y = y x and solve.
x 3 4 x = ( 3 4 x ) x x^{\frac34x} = \left(\frac34x\right)^x x 4 3 x = ( 4 3 x ) x
x 3 4 x = ( 3 4 ) x ⋅ x x x^{\frac34x} = \left(\frac34\right)^x \cdot x^x x 4 3 x = ( 4 3 ) x ⋅ x x
x − 1 4 x = ( 3 4 ) x x^{-\frac14x} = \left(\frac34\right)^x x − 4 1 x = ( 4 3 ) x
x − 1 4 = 3 4 x^{-\frac14} = \frac34 x − 4 1 = 4 3
x = 256 81 x = \frac{256}{81} x = 81 256
y = 3 4 x = 192 81 y = \frac34x = \frac{192}{81} y = 4 3 x = 81 192
x + y = 448 81 x + y = \frac{448}{81} x + y = 81 448
448 + 81 = 529 448 + 81 = \boxed{529} 448 + 81 = 529
Solution 2
We solve in general using c c c instead of 3 / 4 3/4 3/4 . Substituting y = c x y = cx y = c x , we have:
x c x = ( c x ) x ⟹ ( x x ) c = c x ⋅ x x x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x x c x = ( c x ) x ⟹ ( x x ) c = c x ⋅ x x
Dividing by x x x^x x x , we get ( x x ) c − 1 = c x (x^x)^{c - 1} = c^x ( x x ) c − 1 = c x .
Taking the x x x th root, x c − 1 = c x^{c - 1} = c x c − 1 = c , or x = c 1 / ( c − 1 ) x = c^{1/(c - 1)} x = c 1/ ( c − 1 ) .
In the case c = 3 4 c = \frac34 c = 4 3 , x = ( 3 4 ) − 4 = ( 4 3 ) 4 = 256 81 x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81} x = ( 4 3 ) − 4 = ( 3 4 ) 4 = 81 256 , y = 64 27 y = \frac {64}{27} y = 27 64 , x + y = 256 + 192 81 = 448 81 x + y = \frac {256 + 192}{81} = \frac {448}{81} x + y = 81 256 + 192 = 81 448 , yielding an answer of 448 + 81 = 529 448 + 81 = \boxed{529} 448 + 81 = 529 .
Solution 3
Taking the logarithm base x x x of both sides, we arrive with:
y = log x y x ⟹ y x = log x y = log x 3 4 x = 3 4 y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \frac{3}{4}x = \frac{3}{4} y = log x y x ⟹ x y = log x y = log x 4 3 x = 4 3
Where the last two simplifications were made since y = 3 4 x y = \frac{3}{4}x y = 4 3 x . Then,
x 3 4 = 3 4 x ⟹ x 1 4 = 4 3 ⟹ x = ( 4 3 ) 4 x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4 x 4 3 = 4 3 x ⟹ x 4 1 = 3 4 ⟹ x = ( 3 4 ) 4
Then, y = ( 4 3 ) 3 y = \left(\frac{4}{3}\right)^3 y = ( 3 4 ) 3 , and thus:
x + y = ( 4 3 ) 3 ( 4 3 + 1 ) = 448 81 ⟹ 448 + 81 = 529 x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529} x + y = ( 3 4 ) 3 ( 3 4 + 1 ) = 81 448 ⟹ 448 + 81 = 529
Solution 4 (another version of Solution 3)
Taking the logarithm base x x x of both sides, we arrive with:
y = log x y x ⟹ y x = log x y = log x ( 3 4 x ) = 3 4 y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4} y = log x y x ⟹ x y = log x y = log x ( 4 3 x ) = 4 3
Now we proceed by the logarithm rule log ( a b ) = log a + log b \log(ab)=\log a + \log b log ( ab ) = log a + log b . The equation becomes:
log x 3 4 + log x x = 3 4 \log_x \frac{3}{4} + \log_x x = \frac{3}{4} log x 4 3 + log x x = 4 3
⟺ log x 3 4 + 1 = 3 4 \Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4} ⟺ log x 4 3 + 1 = 4 3
⟺ log x 3 4 = − 1 4 \Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4} ⟺ log x 4 3 = − 4 1
⟺ x − 1 4 = 3 4 \Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4} ⟺ x − 4 1 = 4 3
⟺ 1 x 1 4 = 3 4 \Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4} ⟺ x 4 1 1 = 4 3
⟺ x 1 4 = 4 3 \Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3} ⟺ x 4 1 = 3 4
⟺ x 4 = 4 3 \Longleftrightarrow \sqrt[4]{x} = \frac{4}{3} ⟺ 4 x = 3 4
⟺ x = ( 4 3 ) 4 = 256 81 \Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81} ⟺ x = ( 3 4 ) 4 = 81 256
Then find y y y as in solution 3, and we get 529 \boxed{529} 529 .