How do you evaluate #log_3(1/27)#?

1 Answer
Özgür Özer
Dec 19, 2015

#color(white)(xx)-3#

Explanation:

#color(white)(xx)log_3 (1/27)=log_3 (1/3^3)#
#color(white)(xxxxxxxxx)=log_3 3^-3#
#color(white)(xxxxxxxxx)=-3xxlog_3 3#
(The logarithm of the power of a number rule)

#color(white)(xxxxxxxxx)=-3xx1#
#color(white)(xxxxxxxxx)=-3#

Related questions
See all questions in Functions with Base b
Impact of this question
16960 views around the world
You can reuse this answer
Creative Commons License