First note that log(x) is defined only for x > 0; log(x + 3) for x > -3; and log(x - 1) for x > 1. So if we find any solutions x ≤ 1, these will be extraneous.
Condense the right side:
log(x) = log((x + 3) / (x - 1))
Let b be the base of the logarithm. Then writing both sides as powers of this b simplifies the equation to
b^log(x) = b^log((x + 3) / (x - 1))
x = (x + 3) / (x - 1)
Solve for x :
x (x - 1) = x + 3
x ² - x = x + 3
x ² - 2x - 3 = 0
(x - 3) (x + 1) = 0
x - 3 = 0 or x + 1 = 0
x = 3 or x = -1
We throw out x = -1 because it's less than 1, so the only (real) solution is x = 3.
