logarithmic average with uniform weights = geometric mean?

Question

In the paper "BLEU:a Method for Automatic Evaluation of Machine Translation"-Papineni it says at the bottom of the 4th page:

"logarithmic average with uniform weights is equivalent to geometric mean [..]".

Why is that?

Answer 1

Because

1. $\log a + \log b = \log (ab)$
2. $x\cdot \log a = \log(a^x)$

Therefore,

$$\mathrm{average}(\log(a_1), \log(a_2)\dots,\log(a_n)) =\\=\frac1n\sum_{i=1}^n \log(a_i) = \frac1n\log\left(\prod_{i=1}^na_i\right) = \log\left(\left(\prod_{i=1}^n a_i\right)^\frac1n\right)=\log\left(\sqrt[n]{\prod_{i=1}^n a_i}\right) =\\=\log(\mathrm{geometric\_mean}(a_1,a_2,\dots,a_n)$$