How do you solve #Log_3 (x+1) = 2 + Log_3 (x-3)#?

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Answer 1 · 2016-05-04T18:34:02

#x=7/2#

#log_3 (x+1)-log_3(x-3)=2#

#log_3 ((x+1)/(x-3))=2#

#3^2 = (x+1)/(x-3)#

#9=(x+1)/(x-3)#

#9(x-3)=x+1#

#9x-27=x+1#

#9x-x=1+27#

#8x=28#

#x=28/8=7/2#