How do you solve the equation #log_4a+log_4 9=log_4 27#?

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Answer 1 · 2016-10-27T19:32:35

#a=3#

#log_4 a + log_4 9 = log_4 27#

#log_4 a + log_4 9 - log_4 27=0#

#log_4( (9a)/27)=0#----->Use properties #log_b x+log_b y = log_b (xy)# and #log_b x-log_b y = log_b (x/y)#

#4^0=(9a)/27#

#1=(9a)/27#

#27=9a#

#a = 3#