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## Homework Statement

[tex]xln(2x+1)-x+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)-x[/tex]

## Homework Equations

[tex]ln(x^a) = aln(x), ln(xy) = ln(x) + ln(y), ln(\frac{x}{y}) = ln(x) - ln(y)[/tex]

## The Attempt at a Solution

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

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

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