How do you expand #ln(x^2/y^2)#?

1 Answer
Apr 30, 2016

Answer:

#2lnx-2lny#

Explanation:

The first law of logs you will want to use is #ln(a/b)=lna-lnb#, so

#ln(x^2/y^2)=lnx^2-lny^2#

Another law of logs is that an exponent inside the log is the same as a coefficient outside, or #lnc^2=2lnc#, so

#lnx^2-lny^2=2lnx-2lny#