In log-log regression why are the $\beta$1 coefficients the same for non natural logs but the $\beta$0 coefficients different? Can the $\beta$1 coefficients for non natural log-log regression still be interpreted as a 'One percent increase in IV is associated with a $\beta$1 percent increase in DV' Here, it would be for a 1% change in body mass we'd expect to see a 0.6518 % change in brain mass.
body = c(62000,277,5000000,160000,28000,960,200,60000000,37000000,2500)
brain = c(1400,7.5,6000,1700,46.2,7.4,3,6000,7820,12)
fit1 = lm(log(brain)~log(body))
print(fit1)
fit2 = lm(log10(brain)~log10(body))
print(fit2)
fit3 = lm(log2(brain)~log2(body))
print(fit3)
EDIT:
Is the maths for the $log_{10}$ case the following (based on Interpretation of log transformed predictor):
In the log-log- model, see that
$$\begin{equation*}\beta_1 = \frac{\partial \log_{10}(y)}{\partial \log_{10}(x)}.\end{equation*}$$
which gives
$$\begin{equation*} \frac{\partial \log_{10}(y)}{\partial y} = \frac{1}{y} \end{equation*}$$
or
$$\begin{equation*} \partial \log_{10}(y) = \frac{\partial y}{y}. \end{equation*}$$
Best Answer
If you understand how to convert units--such as kilograms to pounds or meters to feet--then you will understand perfectly what is going on here, too, because it involves a simple change of units. (For more about this, please see Gung's answer to How will changing the units of explanatory variables affect a regression model?)
The "base" of a logarithm is its units of measurement. That is, changing bases amounts to multiplication by a constant:
$$\log_b(x) = c\, \log(x)\tag{1}$$
where $c = 1/(\log b)$.
This gives you all you need to answer the questions. Begin with the first model,
$$\log \text{brain} = \alpha + \beta\,\log\text{body}.$$
Both sides use natural logs: that's their common unit of measurement for logarithms. Specifically, $\alpha$ is measured in "nats" and $\beta$ is "nats per nat": it is unitless. Compare this to the second model using $(1)$
$$c \log \text{brain} = \log_{10}\text{brain} = \alpha_{10} + \beta_{10}\,\log_{10}\text{body} = \alpha_{10} + \beta_{10}\,c \log\text{body}.$$
Both sides use common logs. Changing back to the original units by dividing through by $c$ yields
$$\log \text{brain} = \frac{1}{c}\alpha_{10} + \beta_{10}\,\log\text{body}.$$
Comparing this to the first model shows immediately that
$$\alpha = \frac{1}{c}\alpha_{10},\quad \beta = \beta_{10}.$$
In other words,
In particular,
As an example, consider the estimates in your code. It reports $\hat\alpha = -1.8980$ and $\hat\beta = 0.6518$. We therefore anticipate that $\hat\alpha_{10} = (1/c)(-1.8980)$ where $c=\log{10} = 2.30\ldots$. The value works out to $\hat \alpha_{10} = (1/2.30\ldots)(-1.8980) = -0.8243$. Sure enough, that's precisely what the second model outputs for the intercept (and the two slope estimates are the same).