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

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Answer 1 · 2015-10-30T11:37:46

#-8log_(3)8x#

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)