Logarithm equation 
How to solve this logarithm equation?
  $$\frac12\cdot[\log(x) + \log(2)) + \log[\sqrt{2x} + 1] = \log(6).$$

The answer is $2$.
I've tried to solve it, but I don't know how to proceed:
$\frac12\log(2x) + \log[\sqrt{2x} + 1] = \log(6)$
$\log([(2x)^\frac12] + \log[\sqrt{2x} + 1] = \log(6)$ 
$\log[\sqrt{2x}] + \log[\sqrt{2x} + 1] = \log(6)$
$\log(\sqrt{2x}\cdot [\sqrt{2x} + 1]) = \log(6)$
$\sqrt{2x}\cdot[\sqrt{2x} + 1] = 6$
$2x + \sqrt{2x} = 6$
$2x + \sqrt{2}\sqrt{x} - 6 = 0$   

Square both sides of the equation $\sqrt{2x}=6-2x$ to obtain $2x=36-24x+4x^2$ which will give you this quadratic equation $4x^2-26x+36=0$. If we simplify the equation further by dividing by the common factor $2$ we obtain $2x^2-13x+18=0$. This should be easy to solve if you use the quadratic formula or by factorizing the left side and finding its roots.
