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I have the likelihood function $L(\theta) = \prod _{i=1}^n\left(\frac{1}{\theta \:}e^{-\frac{x_i}{\theta \:}}\right)$. I'm trying to take the natural log, $\ln(L(\theta))$, but I'm not sure how this works with respect to $\prod$. Does anyone know what the process for this log is?

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    $\begingroup$ Log of a product is a sum of logs. $\endgroup$
    – Wojowu
    Nov 4 '17 at 22:34
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The log of a product is the sum of logs of the things inside the product. So $$\ln L(\theta)=\sum_{i=1}^n \ln\left(\frac{1}{\theta}e^{-x_i/\theta}\right)=\sum_{i=1}^n \left(\ln\left(\frac{1}{\theta}\right)-\frac{x_i}{\theta}\right)$$

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  • $\begingroup$ I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly. $\endgroup$ Nov 4 '17 at 22:41
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$$\ln{(L(\theta))} =\sum_{i=1} ^n \ln\left(\frac{1}{\theta}e^{-\frac{x_i}{\theta}}\right) $$

This is because of the product rule for logarithms, that says that $\log_a (BC) = \log_a (B) + \log_a (C)$.

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