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If logyx+lo...

Question

If ${log}_{y} x + {log}_{x} y = 7$ then $({log}_{y} x)^{2} + ({log}_{x} y)^{2}$ is equal to

A

43

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B

45

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C

47

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D

49

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Solution

The correct option is C $47$
$l o g_{y} x + l o g_{x} y = 7$ then $(l o g_{y} x)^{2} + (l o g_{x} y)^{2}$
$\to l o g_{y} x + l o g_{x} y = 7$
square both side
$(l o g_{y} x + l o g_{x} y)^{2} = (7)^{2}$
$(l o g_{y} x)^{2} + (l o g_{x} y)^{2} + 2. l o g_{y} x . l o g_{x} y = 49$
$(l o g_{y} x)^{2} + (l o g_{x} y)^{2} + 2. l o g_{y} x . \frac{1}{l o g_{y} x} = 49$
$(l o g_{y} x)^{2} + (l o g_{x} y)^{2} + 2 = 49$
$(l o g_{y} x)^{2} + (l o g_{x} y)^{2} = 47$

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Similar questions

x^{2} + y^{2} - 25 x y, t h e n p r o v e 2 l o g (x + y) = 3 l o g 3 + l o g x + l o g y

I

f (x, y) = \frac{1}{x^{2}} + \frac{1}{x y} + \frac{log x - log y}{x^{2} + y^{2}}

x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}

x^{2} + y^{2} = 25 x y

2 l o g (x + y) = 3 l o g 3 + l o g x + l o g y

x^{2} + y^{2} = 25 x y,

then prove that

2 log (x + y) = 3 log 3 + log x + log y .

x^{2} + y^{2} = 7 x y

2 log (x + y) = log x + log y + 2 log 3

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