How do you solve $log (6 x + 2) - log (3 x - 6) = log 6 - log 2$ $log (6x+2) - log (3x-6) = log 6 - log 2$ ?

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How do you solve $log (6 x + 2) - log (3 x - 6) = log 6 - log 2$ $log (6x+2) - log (3x-6) = log 6 - log 2$ ?

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Answer 1 · 2016-01-28T13:31:14

20/3

Explanation:

$log (6 x + 2) - log (3 x - 6) = log 6 - log 2$ $log(6x+2)-log(3x-6)=log6-log2$
$or, log (\frac{6 x + 2}{3 x - 6}) = log (\frac{6}{2})$ $or,log((6x+2)/(3x-6))=log(6/2)$
$or, \frac{6 x + 2}{3 x - 6} = 3$ $or,(6x+2)/(3x-6)=3$ [by, adding antilog to each side]

$or, 6 x + 2 = 9 x - 18$ $or,6x+2=9x-18$

$or, 3 x = 20$ $or,3x=20$
$or, x = \frac{20}{3}$ $or,x=20/3$

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How do you solve $log (6 x + 2) - log (3 x - 6) = log 6 - log 2$ $log (6x+2) - log (3x-6) = log 6 - log 2$ ?

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How do you solve log(6x+2)−log(3x−6)=log6−log2log (6x+2) - log (3x-6) = log 6 - log 2?

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How do you solve $log (6 x + 2) - log (3 x - 6) = log 6 - log 2$ $log (6x+2) - log (3x-6) = log 6 - log 2$ ?