Question

2log(2y-3x)=logx+logy then x/y=?

Answer 1

Hi ,

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

log ( 2y - 3x )² = log ( xy )

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Since ,

1 ) n log a = log aⁿ

2 ) log m + log n = log mn

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( 2y - 3x )² = xy

( 2y )² + ( 3x )² - 2 × 2y × 3x = xy

4y² + 9x² - 12xy = xy

4y² + 9x² = 13xy

divide each term with xy ,

we get,

4y²/xy + 9x²/xy = 13

4y/x + 9x/y = 13

4/( x/y ) + 9x/y = 13

Let a = x/y ---( 1 )

4/a + 9a = 13

4 + 9a² = 13a

9a² - 13a + 4 = 0

9a² - 9a - 4a + 4 = 0

9a ( a - 1 ) - 4 ( a - 1 ) = 0

( a - 1 ) ( 9a - 4 ) = 0

Therefore ,

a - 1 = 0 or 9a - 4 = 0

a = 1 or a = 4/9

x/y = 1 or x/a = 4/9 [ from ( 1 ) ]

I hope this helps you.

: )





Therefore ,

x/y = ( 2y - 3x )²

I hope this helps you.

: )

Answer 2

2log (2y - 3x) = log x  + log y 

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

log (2xy- 3x) = log xy 

(2y - 3x) = xy 

4y+ 9x -12xy = xy 

4y - 13xy + 9x = 0 

4y - (9 + 4)xy + 9x =0 

4y - 9xy - 4xy + 9x = 0 

y(4y - 9x) -x(4y - 9x) = 0 

(y - x) (4y - 9x) = 0 

By Zero product rule,

x = y  &   4y = 9x 

x/y = 1   Or  4/9 = x/y 




i hope this will help you


-by ABHAY