First note that any solution must belong to all of the intervals
x + 2 > 0 ⇒ x > -2
x > 0
2x - 1 > 0 ⇒ x > 1/2
3x - 12 > 0 ⇒ x > 4
so x must be larger than 4.
Recall that log(a) - log(b) = log(a/b). So we can rewrite
log(x + 2) - log(x) = log(2x - 1) - log(3x - 12)
log((x + 2)/x) = log((2x - 1)/(3x - 12))
Take the antilogarithm/exponential of both sides (note that the base of the logarithm doesn't matter here):
exp[log((x + 2)/x)] = exp[log((2x - 1)/(3x - 12))]
(x + 2)/x = (2x - 1)/(3x - 12)
Eliminate the fractions:
x (3x - 12) • (x + 2)/x = x (3x - 12) • (2x - 1)/(3x - 12)
(3x - 12) (x + 2) = x (2x - 1)
Expand both sides, collect terms and factorize to solve for x :
3x² - 6x - 24 = 2x² - x
x² - 5x - 24 = 0
(x - 8) (x + 3) = 0
x - 8 = 0 or x + 3 = 0
x = 8 or x = -3