I rewrite here the question to avoid visualization problems. Does the following hold (for $a_i \in \mathbb R$ and $a_i>0$)?
$$e^{\sum_i a_i \log(a_i)}=\prod_i a_i^{a_i}$$
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2$\begingroup$ You need the $a_i>0$ for the left hand side to make sense. $\endgroup$– JP McCarthyCommented Mar 16, 2018 at 15:33
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$\begingroup$ Thanks @JpMcCarthy, I'll add it to the question text! $\endgroup$– Lo ScrondoCommented Mar 16, 2018 at 15:36
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2$\begingroup$ If $\log()$ means $\ln()$ i.e. $\log_e()$ then it is true, else not. $\endgroup$– Sujit BhattacharyyaCommented Mar 16, 2018 at 15:40
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2$\begingroup$ I don't see any reason why it should not. Isn't it just a plain expression simplification? $\endgroup$– learnerCommented Mar 16, 2018 at 15:40
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1 Answer
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To sum up what was written in the comments, assuming $a_i>0,\,\forall i$, $$\begin{align}\exp\left(\sum_ia_i\log_b(a_i)\right)&=\exp(a_1\log_b(a_1)+a_2\log_b(a_2)\cdots)\\&=\exp(a_1\log_b a_1)\times\exp(a_2\log_b a_2)\times\cdots\\&=\prod_i\exp(a_i\log_b a_i)\\&=\prod_i \exp\left(a_i\cdot\frac{\ln a_i}{\ln b}\right)\\&=\prod_i {a_i}^{a_i/(\ln b)}\end{align}$$ So this only equals to $\prod_i {a_i}^{a_i}$ if $\ln b=1\iff b=e$.
