How do you solve #\log _ { 2} ( x - 2) - \log _ { 2} ( 2x - 3) = \log _ { 2} 2#?

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Answer 1 · 2017-11-15T20:17:04

#x=4/3#

#log(a)-log(b)=log(a/b)#

So:

#log_2((x-2)/(2x-3))=log_2(2)#

#:.#

#(x-2)/(2x-3)=2#

#->x-2=4x-6#

#3x=4=>x=4/3#