Consider the plot below:
This looks like an exponentially decreasing graph, but it is often misleading to make this judgement by eye. (This plot is in fact $y=e^{-2x}$.) By mapping the axes to a log scale we can identify whether or not this is in fact an exponential. If this is an exponential, its log-log plot will look like this:
where $u=\log(y)$ and $v=\log(x)$.
Now I'm getting confused about how to verify this. As far as my understanding goes, for a function $y=f(x)$, to get the log-log plot we would need to do $\log(y)=f(\log(x))$, as opposed to $\log(y)=\log(f(x)).$ (Is this the correct way to convert to a log-log plot?)
My confusion is stemming from trying to get the log-log plot using the method that I think is incorrect:
\begin{equation}
\begin{split}
\log(y) & = \log(e^{-2x}) \\
& = -2x \\
\implies u & = -2e^v,
\end{split}
\end{equation}
which is in fact the log-log plot shown above. Alternatively, if I use the method I think is correct, $\log(y)=f(\log(x))$, I just don't seem to get anything fruitful:
\begin{equation}
\begin{split}
\log(y) & = e^{-2\log(x)} = \frac{1}{x^2}\\
\implies u & = e^{-2v}
\end{split}
\end{equation}
which isn't the correct graph.
My conclusion/question
The results above force me to conclude that converting $y=f(x)$ from a linear-linear plot to a log-log plot is done by doing $$\log(y)=\log(f(x)),$$ as opposed to my original intuition, which is by doing $$\log(y) = f(\log(x)).$$
Could someone please clear up the correct intuition behind this? I can't understand how my forced conclusion is correct.
Best Answer
Log - log transformations are used to convert power functions to linear plots.
A log-log transformation of $y = ax^b$ would be log $y =\ $log$ \ a + b \ $log$\ x$
So a log-log transformation of the exponential function $y = e^{-2x}$ would become ln $y = -2x$. I don't see what anyone would gain from this.
Just transforming the y values ln$\ y = e^{-2x}$ would linearize the plot $(y = -2x)$ and confirm it as an exponential function.