# What does the slope of a log-log plot represent?

I have been asked to plot the relationship of $log(I)$ versus $log(R)$, where $I$ is the moment of inertia of an object and $R$ is the radius used to calculate $I$. What does the slope of this plot represent?

Also, what would the vertical axis intercept of this plot represent?

$$y = ax^k$$ taking logs we have $$\log y = k \log x + \log a$$ if we relabel as $$\bar{y} = k \bar{x} + c$$ we should see that the gradient of the last equation i.e. the $k$, maps to be the gradient in the log-log plot which in turn maps to being the exponent of the original equation.
The Intercept $c = \log a$ which is basically the coefficient of the power law.
• So, if $I$ is proportional to $R^2$, then the slope of the log-log plot should be 2? – Alexander Nov 17 '16 at 17:16
• yes one should conclude that if you plot $\log I$ on the $y$-axis and $\log R$ on the other then yes the gradient will be $2$. – Chinny84 Nov 17 '16 at 17:20
• Thank you! Sorry if this is getting more off topic/is a stupid question, but would the slope of an $I$ versus $R^2$ plot represent the $k$ value in $I = kR^2$? – Alexander Nov 17 '16 at 17:27
• Why would $I = k$ when $R=0$? Since they are multiplied, should $I$ not equal 0? Also, I did not mean to use the same variable. So, if $k$ does not function as the coefficient of the power law, would the slope of $I=kR^2$ equal $k$? – Alexander Nov 17 '16 at 17:35