Question

If 4x^2 +y^2=20xy,then prove that 2log(2x-y)=4log2+logx+logy

Answer 1

Answer:

$2log(2x-y)=4\:log2+logx+logy$

Step-by-step explanation:

Formula used:

Product rule:

$log(MNP)=logM+logN+logP$

Power rule:

$logM^n=n\:logM$

Given:

$4x^2+y^2=20xy\\\\4x^2+y^2=4xy+16xy\\\\4x^2+y^2-4xy=16xy\\\\(2x-y)^2=16xy$

Taking logarithm on bothsides

$log(2x-y)^2=log16xy\\\\log(2x-y)^2=log16+logx+logy\\\\log(2x-y)^2=log2^4+logx+logy\\\\2log(2x-y)=4\:log2+logx+logy$

Answer 2

Answer:

Proved

Step-by-step explanation:

If 4x^2 +y^2=20xy,then prove that 2log(2x-y)=4log2+logx+logy

4x² +y²=20xy

subtracting 4xy from both sides

=> 4x² + y² - 4xy = 20xy - 4xy

=> (2x)² + y² -2(2x)y = 16xy

using (a-b)² = a² + b² - 2ab

here a = 2x , b = y

=> (2x-y)² = 2⁴xy

Taking Log both sides

2Log(2x-y) = 4Log2 + logx + logy

Hence Proved