The Geometric Mean is just the Arithmetic Mean of the Exponents

This one is going to be brief, but I find it interesting. A buddy of mine a few months back sent me a text about this realization: The geometric mean is just the arithmetic mean of the exponents! I'll explain in a second but just a quick background.

An arithmetic mean is what we commonly refer to as the average: add all the elements up and divide by the number of elements. $$\overline{x}=\frac{x_1+x_2+\cdots+x_n}{n}$$ Alternatively, the we have the geometric mean. Except the geometric mean is the multiplicative analog to the arithmetic mean. The geometric mean is the $n$^th root of the product of the elements. $$\overline{y}'=\left(y_1y_2\cdots y_n\right)^{\frac{1}{n}}$$ The connection is as follows. Each number $y_i$ (assuming they are all positive) can be expressed as $y_i=a^{b_i}$ for some $a,b\in\mathbb{R}$. Therefore we have the following $$\begin{align*} \overline{y}'&=\left(y_1y_2\cdots y_n\right)^{\frac{1}{n}}\\ &=\left(a^{b_1}a^{b_2}a^{b_2}\cdots a^{b_n}\right)^{\frac{1}{n}}\\ &=\left(a^{b_1+b_2+\cdots+b_n}\right)^{\frac{1}{n}}\\ &=a^{(b_1+b_2+\cdots+b_n)/n} \end{align*}$$ which states that if each $y_i$ can be expressed as $y_i=a^{b_i}$ for some $a\in\mathbb{R}$ and $b_i\in\mathbb{R}$ then the geometric mean of all $y_i$ is the base, $a$, raised to the arithmetic mean of all exponents $b_i$. $$\overline{y}'=a^{\overline{b}}$$ Pretty interesting, eh?