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$n \ln(\frac{1}{2 \theta^3}) = n \ln (1) - 3n \ln (2\theta)$

why is this not right?

apparently the answer is $-n \ln (2) - 3 n \ln (\theta)$

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  • $\begingroup$ Note that $3n \ln(2\theta) = n \ln(2^3\theta^3)$ $\endgroup$ – Brian Mar 8 at 16:36
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$n\ln(\frac{1}{2{\theta}^3})$ can be written as $n\ln(2^{-1}{\theta}^{-3})$.So according to the rule $$\ln(xy)=\ln{x}+\ln{y}$$, you will get the right answer.

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We have $$n\ln\left(\frac{1}{2\theta^3}\right)=n\left(\ln(1)-\ln(2\theta^3)\right)=n\left(-\ln(2)-3\ln(\theta)\right)$$

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  • $\begingroup$ is this equivalent tho? : "$n \ln (1) - 3n \ln (2\theta)$" $\endgroup$ – Tinler Mar 8 at 16:33
  • $\begingroup$ Yes since $$n\ln(1)-3n\ln(2\theta)=-3n\ln(2\theta)=-3n(\ln(2)+\ln(\theta))$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 8 at 16:35
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The exponent $3$ is only on $\theta$, not on the $2$, so you can't pull it out like that. You have to split off the $2$ first:

$$\ln(2\theta^3) = \ln 2 + \ln \theta^3 = \ln 2+3\ln \theta.$$

Next, note that $\ln 1 = 0.$

So you should have

$$n\ln \frac{1}{2\theta^3} = n\ln 1 - n\ln(2\theta^3)$$

and then plug in the above.

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