How do you evaluate #log_4 80 - log_4 5#?

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Answer 1 · 2016-05-28T08:50:51

#log _4 80-log_4 5=2#

#log _4 80-log_4 5=log _4 (80/5)#

#log _4 80-log_4 5=log_4 16#

#log _4 80-log_4 5=log_4 4^2#

#log _4 80-log_4 5=2*log_4 4#

#log_4 4=1#

#log _4 80-log_4 5=2*1#

#log _4 80-log_4 5=2#