If $(3 x)^{l o g 3} = (4 y)^{l o g 4}$ and $(4)^{l o g x} = (3)^{l o g y}$ , then x equals

1/3

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1/2

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1/4

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Solution

The correct option is A
1/3

$(3 x)^{l o g 3} = (4 y)^{l o g 4} l o g 3 l o g_{3} 3 x = l o g 4 l o g_{3} 4 y l o g_{4} 3 (l o g_{3} 3 + l o g_{3} x) = l o g_{3} 4 + l o g_{3} y \to (1) 4^{l o g x} = 3^{l o g y} l o g x . l o g 4 = l o g y . l o g 3 l o g_{3} x = l o g_{4} y \to (2) (2) i n (1) l o g_{4} 3 (1 + l o g_{4} y) = l o g_{3} 4 + l o g_{4} y l o g_{3} 4 l o g_{4} 3 (1 + l o g_{4} y) = l o g_{3} 4 (1 + l o g_{4} y) (l o g_{4} 3 - l o g_{3} 4) (1 + l o g_{4} y) = 0 l o g_{4} y = - 1 = l o g_{3} x x = 3^{- 1} x = \frac{1}{3}$

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Similar questions

If $(3 x)^{l o g 3} = (4 y)^{l o g 4}$ and $(4)^{l o g x} = (3)^{l o g y}$ , then x equals

(3 x)^{log 3} = (4 y)^{log 4}

4^{log x} = 3^{log y}

x

(3 x)^{log 3} = (4 y)^{log 4}

4^{log x} = 3^{log y}

x a n d y

(x, y)

(3 x)^{log 3} = (4 y)^{log 4}, 4^{log x} = 3^{log y}

2 log x + 3 log y = log 4

x^{2} y^{3} =

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