Help Please!!!!!

Domain: {x element R : x>0} (all positive real numbers)
(assuming a function from reals to reals)

Solve for x:
(log(2 x^2+3 x))/(log(10)) = (log(6 x+2))/(log(10))

Subtract (log(6 x+2))/(log(10)) from both sides:
(log(2 x^2+3 x))/(log(10))-(log(6 x+2))/(log(10)) = 0

Bring (log(2 x^2+3 x))/(log(10))-(log(6 x+2))/(log(10)) together using the common denominator log(10):
-(log(6 x+2)-log(2 x^2+3 x))/(log(10)) = 0

Multiply both sides by -log(10):
log(6 x+2)-log(2 x^2+3 x) = 0

log(6 x+2)-log(2 x^2+3 x) = log(6 x+2)+log(1/(2 x^2+3 x)) = log((6 x+2)/(2 x^2+3 x)):
log((6 x+2)/(2 x^2+3 x)) = 0

Cancel logarithms by taking exp of both sides:
(6 x+2)/(2 x^2+3 x) = 1

Multiply both sides by 2 x^2+3 x:
6 x+2 = 2 x^2+3 x

Subtract 2 x^2+3 x from both sides:
-2 x^2+3 x+2 = 0

The left hand side factors into a product with three terms:
-(x-2) (2 x+1) = 0

Multiply both sides by -1:
(x-2) (2 x+1) = 0

Split into two equations:
x-2 = 0 or 2 x+1 = 0

Add 2 to both sides:
x = 2 or 2 x+1 = 0

Subtract 1 from both sides:
x = 2 or 2 x = -1

Divide both sides by 2:
Answer: | x = 2      or     x = -1/2 (assuming a complex-valued logarithm)

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)