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

$rArr 8x - 4 = x + 3$

$rArr 7x = 7$

$rArr x = 1$

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