How do you rewrite #Log_3 8 - 4log_3(x^2)# as a single log?

1 Answer
Oct 30, 2015

Answer:

#-8log_(3)8x#

Explanation:

By using laws of logs and exponents we get:

#log_(3)8-4log_3x^2=log_(3)8-8log_3x#

#=log_(3)8-log_3x^8#

#=log_3(8/x^8)#

#=log_(3)8x^(-8)#

#=-8log_(3)8x#

(Note: There might be other valid ways to express it as well)