How do you solve $log x - log 3 = log (x + 8)$ $logx-log3 = log(x+8)$ ?

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Answer 1 · 2016-01-27T20:21:31

$x = - 12$ $x=-12$

Subtracting logs is the result from taking loges of a division

So $log (x) - log (3) \to log (\frac{x}{3})$ $log(x)-log(3) -> log(x/3)$

Write as $. . log (\frac{x}{3}) = log (x + 8)$ $color(white)(..)log(x/3)=log(x+8)$

If this is true then it is also true that:

$\frac{x}{3} = x + 8$ $x/3=x+8$

$x = 3 x + 24 \to$ $x=3x+24->$ Multiplied both sides by 3

$2 x = - 24 \to$ $2x=-24->$ Collecting like terms and simplifying

$x = - 12 \to$ $x=-12->$ Divide both sides by 2

How do you solve logx-log3 = log(x+8)logx−log3=log(x+8)logx-log3 = log(x+8)?