How do you evaluate #\log _ { 3} x + 2\log _ { 3} x#?
1 Answer
Nov 30, 2017
Explanation:
#"simplify using the "color(blue)"laws of logarithms"#
#•color(white)(x)logx+logyhArrlogxy#
#•color(white)(x)logx^nhArrnlogx#
#rArrlog_3x+log_3x^2=log_3x^3#
