Natural Logarithm Manupulations

  • Thread starter bob1182006
  • Start date
  • #1
492
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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]
 
Last edited:

Answers and Replies

  • #2
682
1
factor out the ln(2x+1)
 
  • #3
492
1
factor out the ln(2x+1)
:surprised wow I can't believe I didn't see that, thanks so much I get it now
 

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