Question

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

Answer 1

Given 4x^2+y^2=20xy

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

(2x-y)^2=(2^4)xy

apply log on both sides

log(2x-y)^2=log(2^4)xy

[as logm^n= nlogm and logmn=logn+logm]

2log(2x-y)=log2^4+logx+logy

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

Hence proved

Answer 2

Answer:

4x² + y² = 20xy

=> 4x² - 4xy + y² = 16xy

=> (2x)² - 2(2x)y + y² = 2⁴xy

=> ( 2x - y )² = 2⁴xy

=> log [ ( 2x - y )² ] = log (2⁴xy)

=> 2 log ( 2x - y ) = log 2⁴ + log x + log y

=> 2 log ( 2x - y ) = 4 log 2 + log x + log y