How do you condense #log(pi) + 2log r#?

1 Answer
Jun 1, 2016

Answer:

#log(pi)+2log(r)=color(green)(log(pir^2)#

Explanation:

Remember
#color(white)("XXX")a*log(b) = log(b^a)#
and
#color(white)("XXX")log(c)+log(d)=log(c*d)#

Therefore
#log(pi)+2log(r)#
#color(white)("XXX")=log(pi)+log(r^2)#

#color(white)("XXX")=log(pir^2)#