"Linearize" an exponential-looking graph with log function

Suppose $y=Ae^{kx}$. Then, taking logs, we see that $\log y = kx + \log A$.

Therefore, try taking logs of the $y$-values, and plot those against the ordinary $x$-values. You'll get a straight line whose slope tells you $k$ and whose $y$-intercept tells you $\log A$.