How do you condense #log_3 (9xy^2)-log_3 (27xy)#?

1 Answer
Jul 9, 2016

Answer:

#log_3(1/3 y)#

Explanation:

Using the #color(blue)"law of logarithms"#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(loga-logb=log(a/b))color(white)(a/a)|)))#

#rArrlog_3(9xy^2)-log_3(27xy)=log_3((9xy^2)/(27xy))#

which then simplifies to #log_3(y/3)#