# Natural Logarithm Manupulations

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

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