Question

4x² + y² =20xy
then show
that through Log
(2x-y = 4 log2 + logx + log y )​

Answer 1

Answer:

Answer:

2log(2x-y)=4\:log2+logx+logy2log(2x−y)=4log2+logx+logy

Step-by-step explanation:

Formula used:

Product rule:

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

Power rule:

logM^n=n\:logMlogMn=nlogM

Given:

\begin{gathered}4x^2+y^2=20xy\\\\4x^2+y^2=4xy+16xy\\\\4x^2+y^2-4xy=16xy\\\\(2x-y)^2=16xy\end{gathered}4x2+y2=20xy4x2+y2=4xy+16xy4x2+y2−4xy=16xy(2x−y)2=16xy

Taking logarithm on bothsides

\begin{gathered}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\end{gathered}log(2x−y)2=log16xylog(2x−y)2=log16+logx+logylog(2x−y)2=log24+logx+logy2log(2x−y)=4log2+logx+logy