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I know that $\sum_i \log(x_i) = \log(\prod_ix_i)$. I am wondering if the identity extends to the case where we have $\sum_ia_i\log(x_i)$ where $a_i$ are positive weights. If not, is there any way to simplify $\sum_ia_i\log(x_i)$?

Rong Zhang
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1 Answers1

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Note that $$\sum_ia_i\log(x_i)=\sum_i\log(x_i^{a_i})=\log\left(\prod_i x_i^{a_i}\right).$$

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