# When can taking the following transformation of data $log_{10}(e^{mean(log(x))})$ be useful?

I wished to transform some data and have noticed that the geometric mean is a known transformation to normalise data with high differences in comparably large values relative to lower values.

However, has such a transformation been used before: $$log_{10}(e^{mean(log(x))})$$

When does using the $$log_{10}$$ of the geometric mean become meaningful? for example, if I have a set of values and I want to take the differences likes:

x <- data.frame(m = rep(1:2, 2),x=c(46, 7, 888, 9), y=c(888, 6, 77, 5), d=c(999, 66, 5, 4))

#geometric mean
ge <- function(x){
+     exp(mean(log(x)))
+ }

#log10 mean
ge <- function(x){
+     log10(exp(mean(log(x))))
+ }

#output
> aggregate(. ~ m, x, ge)
m          x          y        d
1 1 202.108882 261.488049 70.67531
2 2   7.937254   5.477226 16.24808
> aggregate(. ~ m, x, log.ge)
m         x         y        d
1 1 2.3055854 2.4174518 1.849268
2 2 0.8996703 0.7385606 1.210802


If I were to take the differences from the rows of 2 from 1, by ratio it is far larger in the geometric mean in comparison to taking the $$log_{10}$$ how much does this impact the data?

• $\log_{10}(\exp(x))$ is just the same as multiplying $x$ by a constant - specifically by $1/\ln(10)$. So your expression is just $k.\text{mean}(\log(x))$ for $k\approx 0.43$. Jul 14, 2021 at 12:06
• @Glen_b Competitively speaking - are there any charges by incurring the constant $k$ relative to $exp$ on the data-transformation - such costs may be the size of the data as above. What is your input on the profit of maintaining the constant $k$, any given situation by which this is apprehensively useful? Jul 14, 2021 at 12:18
• Using $\log_{10}$ can be helpful (a) when you do not have a calculator and have to use tables, and (b) when you want to express results as related to powers of $10$ as in for example decibels. Jul 14, 2021 at 12:56
• @Stackcans I am sorry, I really don't follow what you're asking. Jul 15, 2021 at 5:58

Introducing the log is equivalent to scaling the (non-exponentiated) mean by $$\frac{1}{\log_e(10)}$$. Notice that you can change the base of a logarithm from $$a$$ to $$b$$ by scaling: \begin{align} \log_a{x} = \frac{\log_b(x)}{\log_b(a)} \end{align} We therefore have: \begin{align} \log_{10}\left(e^{\text{mean}(\log(x))}\right) & = \frac{1}{\log_{e}(10)}\log_e\left(e^{\text{mean}(\log(x))}\right) \\ & = \frac{1}{\log_{e}(10)}\text{mean}(\log(x)) \end{align}