e^x and ln x: a pair of inverses

$e = 2.718281828\ldots$

Like $\pi$, it's a special number whose decimals never end.

It's not an arbitrary choice. The curve $y = e^x$ has a gradient equal to its own height at every point. In words: the rate of growth equals the amount present.

That is why $e$ turns up in models of populations, radioactive decay, cooling and compound interest.

$\ln x$ means $\log_e x$: "the power of $e$ that gives $x$". So everything true of logs is true of $\ln$:

• $\ln 1 = 0$ (because $e^0 = 1$)
• $\ln e = 1$ (because $e^1 = e$)
• $\ln(xy) = \ln x + \ln y$, and the other log laws

$e^x$ and $\ln x$ undo each other:

$e^{\ln x} = x \;\;(x > 0) \qquad\qquad \ln(e^x) = x \;\;(\text{all } x)$

A function followed by its inverse returns the original input. That is what "inverse" means.

On a graph, inverse functions are reflections in the line $y = x$. For example, the point $(0, 1)$ on $y = e^x$ becomes $(1, 0)$ on $y = \ln x$.

(An asymptote is a line the graph gets closer and closer to but never touches.)

The domain and range swap over, exactly as they must for an inverse pair. That's why $\ln(\text{negative})$ doesn't exist.

Here $k$ is a constant: a fixed number like $2$ or $-0.5$.

• $k > 0$: growth. The curve rises, passing through $(0, 1)$ and heading up steeply. Bigger $k$ means faster growth. More negative $k$ means faster decay.
• $k < 0$: decay. The curve falls towards the $x$-axis, which stays as an asymptote.
• Both pass through $(0, 1)$, because $e^0 = 1$ whatever $k$ is.
• Neither ever touches the $x$-axis: $e^{kx}$ is always positive.

Why is decay still positive? Write $k = -c$ with $c$ positive. Then

$e^{-cx} = \frac{1}{e^{cx}}$

That is 1 divided by a positive number, so it is positive too.