# rewrite $n \ln(\frac{1}{2 \theta^3})$

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

why is this not right?

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

• Note that $3n \ln(2\theta) = n \ln(2^3\theta^3)$ – Brian Mar 8 at 16:36

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

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

• is this equivalent tho? : "$n \ln (1) - 3n \ln (2\theta)$" – Tinler Mar 8 at 16:33
• Yes since $$n\ln(1)-3n\ln(2\theta)=-3n\ln(2\theta)=-3n(\ln(2)+\ln(\theta))$$ – Dr. Sonnhard Graubner Mar 8 at 16:35

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.