Question

4 x square + 9 y square is equal to 16xy then show that 2 log(2x-3y)=2 log2+ logx+ logy​

Answer 1

Answer:

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Answer 2

Hence proved.

Step-by-step explanation:

Given:

$= > 4 {x}^{2} + 9 {y}^{2} = 16xy$

$= > {(2x)}^{2} + {(3y)}^{2} - 16xy = 0$

$= > {(2x)}^{2} + {(3y)}^{2} - 2 \times 2x \times 3y - 4xy = 0$

$= > {(2x - 3y)}^{2} = 4xy$

On taking Log on both side we get,

$log( {(2x + 3y)}^{2} ) = log(4xy)$

As we know that,

$log( {a}^{n} ) = n log(a)$

$log(xy) = log(x) + log(y)$

$= > 2 log((2x + 3y)) = log(4) + log(2x) + log(3y)$

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

Hence proved.