How do you solve #Log (x + 3) - Log (2x - 1) = Log 4#?

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Answer 1 · 2016-01-07T22:54:52

x = 1

on the left hand side of the equation we make use of the following law of logs :

# logx -log y = log(x/y) #

#rArr log(x + 3 ) - log(2x - 1 ) = log ((x + 3 )/(2x - 1 )) #

#rArr log((x + 3 )/(2x - 1 )) = log4 #

#rArr (x + 3 )/(2x - 1 ) = 4 #

and so 4 ( 2x - 1 ) = x + 3

#rArr 8x - 4 = x + 3 #

#rArr 7x = 7 #

#rArr x = 1#