How do you solve #log_2 (x+1) - log_2 (x-1) = 4#?

1 Answer
Alan P.
Dec 27, 2015

#x=17/15#

Explanation:

#log_2(x+1)-log_2(x-1) = log_2((x+1)/(x-1))#

#2^4 = 16 rArr 4=log_2(16)#

Therefore
#color(white)("XXX")log_2(x+1)-log_2(x-1)=4#
is equivalent to
#color(white)("XXX")log_2((x+1)/(x-1))=log_2(16)#

#color(white)("XXX")(x+1)/(x-1) = 16#

#color(white)("XXX")x+1 = 16x-16#

#color(white)("XXX")-15x = -17#

#color(white)("XXX")x=17/15#

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