# What is the correct way to convert between linear-linear and log-log plots?

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.

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.
• OK, so obtaining a log-log plot is done with $\log(y)=\log(f(x))$, thanks. By the way, it's useful to do a log-log when want to check that a heavy-tailed graph is in fact heavy-tailed, as opposed to an exponential in disguise. The heavy-tailed log-log is linear, whereas the exponential log-log isn't. – quanty Jun 8 '18 at 7:56