How do you solve # log (2x + 2) – log (x + 6) =0#?

1 Answer
Alan P.
Mar 7, 2018

#x=4#

Explanation:

Note 1:
#color(white)("XXX")log(A)-log(B)=log(A/B)#
Note 2:
#color(white)("XXX")log(1)=0#

Therefore
#color(white)("XXX")log(2x+2)-log(x+6)=0#
can be written as
#color(white)("XXX")log((2x+2)/(x+6))=log(1)#
which implies
#color(white)("XXX")(2x+2)/(x+6)=1#

#color(white)("XXX")rArr 2x+2=x+6#

#color(white)("XXX")rArr x=4#

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