Natural Logarithm Manupulations

[bob1182006]

Homework Statement

$$xln(2x+1)-x+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)-x$$

Homework Equations

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

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 $xln(2x+1)+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)$

 

[tim_lou]

factor out the ln(2x+1)

 

[bob1182006]

factor out the ln(2x+1)

wow I can't believe I didn't see that, thanks so much I get it now