How do you evaluate #4\log ( x ^ { 3} y ^ { 2} ) + 8\log ( \frac { 1} { x } ) - 4\log ( y )#?

1 Answer
Shwetank Mauria
Dec 4, 2017

#4log(x^3y^2)+8log(1/x)-4logy=4logx+4logy=logx^4y^4#

Explanation:

Using logarithmic formulas #logab=loga+logb#, #log(a/b)=loga-logb#, #loga^m=mloga#, #log(1/a)=-loga#, we can evaluate #4log(x^3y^2)+8log(1/x)-4logy# as follows:

#4log(x^3y^2)+8log(1/x)-4logy#

= #4(logx^3+logy^2)+8(-logx)-4logy#

= #4(3logx+2logy)-8logx-4logy#

= #12logx+8logy-8logx-4logy#

= #4logx+4logy#

= #logx^4+logy^4#

= #logx^4y^4#

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