Solve logarithmic equation I'm getting stuck trying to solve this logarithmic equation:
$$
 \log( \sqrt{4-x} ) - \log( \sqrt{x+3} ) = \log(x)
$$
I understand that the first and second terms can be combined & the logarithms share the same base so one-to-one properties apply and I get to:
$$
 x = \frac{\sqrt{4-x}}{ \sqrt{x+3} }
$$
Now if I square both sides to remove the radicals:
$$
 x^2 = \frac{4-x}{x+3}
$$
Then:
$$
 x^2(x+3) = 4-x
$$
$$
 x^3 +3x^2 + x - 4 = 0
$$
Is this correct so far? How do I solve for x from here?

It is correct so far.  
There is clearly a root between $0$ and $1$.  Either use numerical methods to find it is about $0.893289$ or (not recommended) solve the cubic to get $$\sqrt[3]{\frac{3}{2} - \sqrt{\frac{211}{108}}}  + \sqrt[3]{\frac{3}{2} + \sqrt{\frac{211}{108}}} -1$$ 
