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$$ \left(\frac25\right)\ln(1/2)+\left(\frac15\right)\ln(2) $$

Im having some difficulty with above quotient. Here is what i try to do.

$$ \left(\frac15\right)(2\ln(1/2)+\ln(2)) $$

$$ \left(\frac15\right)(\ln((1/2)^2)+\ln(2)) $$

$$ \left(\frac15\right)(\ln(1/4)+\ln(2)) $$

$$ \left(\frac15\right)(\ln(1)-\ln(4)+\ln(2)) $$

$$ \left(\frac15\right)(\ln(2)-(\ln(4)) $$

$$ \left(\frac15\right)(\ln(1/2)) $$

$$ \left(\frac15\right)(\ln(1)-\ln(2)) $$

Wolfram gets, but im not sure how $$ -3\log(2)/5 $$

Pierre
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    Given the distinction made, what base logarithm is $\log$ here ? Also - you're missing a bracket second line up from the bottom. – pshmath0 Mar 13 '14 at 13:12
  • are both logarithms in $\left(\frac25\right)log(1/2)+\left(\frac15\right)ln(2)$ to the same base? – Guy Mar 13 '14 at 13:13
  • yes they are the same, im used to using ln. – Pierre Mar 13 '14 at 13:14
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    Much quicker to start with $\log(1/2)=-\log 2$. – David Mitra Mar 13 '14 at 13:15
  • @DavidMitra yes. He writes things like $\log(1/2)=\log(1)-\log(2)=-\log(2)$ I am not sure why? – Guy Mar 13 '14 at 13:16
  • @DavidMitra et al. I'm sure the OP is just learning rules of logarithm, including the log of a quotient $\ln(\frac xy) = \ln x - \ln y$. I'm sure s/he will "get" in no time that $\ln(\frac 1x) = -\ln x$. – amWhy Mar 13 '14 at 13:19
  • Pierre, your work is just fine! Just simplify your "ending" answer: $\dfrac{\ln(1) - \ln (2)}{5} = \dfrac {-\ln 2}{5}$, that's all. – amWhy Mar 13 '14 at 13:23

1 Answers1

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You must have mistyped the question into wolfram. Wolfram gets the same answer as you do.

Guy
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