How do you solve $log (2 x + 4) = log (3 x - 12)$ $log (2x + 4) = log (3x -12)$ ?

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Answer 1 · 2018-03-06T11:28:21

$x = 16$ $x=16$

$log (2 x + 4) = log (3 x - 12)$ $log(2x+4) = log(3x-12)$

Remember that: $log a = log b \to a = b$ $loga=logb ->a=b$ (Assuming the same base)

Hence, in this example: $2 x + 4 = 3 x - 12$ $2x+4 = 3x-12$

$3 x - 2 x = 4 + 12$ $3x-2x = 4+12$

$x = 16$ $x = 16$

How do you solve log(2x+4)=log(3x−12)log (2x + 4) = log (3x -12)?