Natural Logarithm Manupulations

1. The problem statement, all variables and given/known data
$$xln(2x+1)-x+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)-x$$

2. Relevant equations
$$ln(x^a) = aln(x), ln(xy) = ln(x) + ln(y), ln(\frac{x}{y}) = ln(x) - ln(y)$$

3. The attempt at a solution

I have no idea how you can go from $xln(2x+1)-x+\frac{1}{2}ln(2x+1)$ to $\frac{1}{2}(2x+1)ln(2x+1)-x$ could someone point me in the right direction?

I know both sides have the -x term, so the only change takes place in $xln(2x+1)+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)$
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 factor out the ln(2x+1)

 Quote by tim_lou factor out the ln(2x+1)
wow I can't believe I didn't see that, thanks so much I get it now