(adsbygoogle = window.adsbygoogle || []).push({}); 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]

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Natural Logarithm Manupulations

**Physics Forums | Science Articles, Homework Help, Discussion**