How do you solve #2log_3y-log_3(y+4)=2#?

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Answer 1 · 2015-11-24T23:54:43

#y=12#

Recall that:

#{(bloga=loga^b),(loga-logb=log(a/b)):}#

#2log_3y-log_3(y+4)=2#

#log_3(y^2)-log_3(y+4)=2#

#log_3(y^2/(y+4))=2#

#y^2/(y+4)=3^2=9#

#y^2=9(y+4)=9y+36#

#y^2-9y-36=0#

#(y-12)(y+3)=0#

#y=12# or #color(red)cancel(y=-3#

(In #loga#, #a>0#.)