log (2x^2 + 3x) = log (6x + 2)
Since the bases are the same, we can solve for the anti-logs
2x^2 + 3x = 6x + 2 rearrange as
2x^2 + 3x - 6x - 2 = 0 simplify
2x^2 - 3x - 2 = 0 factor
(2x + 1) (x - 2) = 0 set each factor to 0 and we get the possible solutions of x = -1/2 or x = 2
Reject x = -1/2 since it would result in taking the log of a negative number on both sides
Then.......x = 2 is the only (real) solution
The domain of the equation on the left would be (-inf, -3/2) U (0, inf)
The domain of the equation on the right would be (-1/3, inf)
Since we must choose the most restrictive interval, the domain for both functions is (0, inf)