Why does a linear line on a logscale vs. logscale plot imply an exponential relashionship? I've seen this type of plot many times in the context of exponential relationships (e.g., Bak's sandhill). If I increase the variable on the x-axis tenfold, and I get back a tenfold increase in y, isn't the relationship between y and x linear?
 A: A linear relation between the logarithm of $x$ and the logarithm of $y$ looks like this:
$$a\log(x) = b\log(y),$$
by properties of the logarithm one can rewrite this to
$$\log(x^a) = \log(y^b)$$
and thus
$$x^a = y^b$$
which is not a linear relation in general but rather a power relation (if that is the correct terminology).
A: A straight line in the $tx$-plane has the equation $t = mx + c$ where $m$ is the gradient and $c$ is the $t$-intercept. Let's say you're plotting time against distance, well, $\log t$ against $\log x$. You might have:
$$\log t = m\log x + c \, . $$
We can use each side of this equation as an exponent:
\begin{array}{ccc}
10^{\log t} &=& 10^{m\log x + c} \\
10^{\log t} &=& \left(10^{\log x}\right)^m \times 10^c \\
t &=& kx^m
\end{array}
In step two I used the two rules that $10^{ab} = (10^a)^b$ and $10^{a+b} = 10^a \times 10^b$. In step three I just relabelled the number $10^c$ as $k$. They are both fixed numbers and $k$ just looks neater. Hence:
$$\log t = m\log x + c \iff t = kx^m$$
You might have seen this written as $t \propto x^m$. That just ignores the constant $k$.
