Natural Logarithm Manupulations

bob1182006

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


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


3. 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]

tim_lou

factor out the ln(2x+1)

bob1182006

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