How do you solve the equation #log_2(y+2)-log_2(y-2)=1#?

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Answer 1 · 2016-10-23T02:12:27

#y=6#

#log_2(y+2)-log_2(y-2)=1#

We can combine the two terms on the left side:

#log_2((y+2)/(y-2))=1#

We can now deal with the log:

#2^(log_2((y+2)/(y-2)))=2^1#

#(y+2)/(y-2)=2#

#(y+2)=2(y-2)#

#y+2=2y-4#

#6=y#

And let's check it (which can also help get more comfortable with log operations):

#log_2(y+2)-log_2(y-2)=1#

#log_2(6+2)-log_2(6-2)=1#

#log_2(8)-log_2(4)=1#

#3-2=1#

#1=1#