Question

If 4xsquare +9ysquare=16 xy, then show that 2 log (2x-3y)=2 log 2+log x+logy​

Answer 1

Answer:

$\bf\,2\,log(2x-3y)=2\,log2+logx+logy$

Step-by-step explanation:

$\text{Given:}$

$4x^2+9y^2=16xy$

$\implies\,4x^2+9y^2=12xy+4xy$

$\implies\,4x^2+9y^2-12xy=4xy$

$\implies\,(2x)^2+(3y)^2-2(2x)(3y)=4xy$

Using, the algebraic identity

$\boxed{\bf\,a^2+b^2-2ab=(a-b)^2}$

$\implies\,(2x-3y)^2=4xy$

$\text{Taking logarithm on both sides}$

$\implies\,log(2x-3y)^2=log4xy$

Using,

$\boxed{\bf\,logMN=logM+logN}$

$\boxed{\bf\,logM^n=n\,logM}$

$\implies\,2\,log(2x-3y)=log4+logxy$

$\implies\,2\,log(2x-3y)=log2^2+logx+logy$

$\implies\,\boxed{\bf\,2\,log(2x-3y)=2\,log2+logx+logy}$

Answer 2

$log4xy = log4xy$

Step-by-step explanation:

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

Proved that,

$2log(2x-3y) = 2log2 + logx + logy$

⇒ $log(2x-3y)^{2} = log(2)^{2} + log{x} + log{y}$

⇒ $log({4x^{2} + 9y^{2} - 12xy}) = log(4xy)$

⇒ $log(16xy-12xy) = log4xy$

⇒ $log4xy = log4xy$

                                              Proved.