Solving for $x$ using exponential log laws For    
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
I moved all the $x$ to left side, used got rid of log and got
$x-(x-1)^2 - (x+1) = 4$
Simplyifing I get
$x^2-2x=4$
The answer should be $x = 4$ (I checked on wolfram alpha)
Help please?

Here are the steps I would use:
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
$\log_2(x) + \log_2\left((x-1)^2\right) = \log_2\left(2^2\right) + \log_2(2x+1)$
That step turns each term into a base $2$ logarithm. Now they can be combined using the product rule for logs:
$\log_2\left(x(x-1)^2\right) = \log_2\left(4(2x+1)\right)$
Now you can get rid of the logarithms, and you get:
$x(x-1)^2 = 4(2x+1)$,
or
$x^3 - 2x^2 - 7x - 4 = 0$
This can be factored as $(x-4)(x + 1)^2=0$, which has solutions $x=4$ and $x=-1$. However, the second of these solutions is extraneous, which you can see by trying to plug it into the original equation.
Does that all makes sense?
