The Natural Logarithm

Idea

There is a second logarithm with a shorthand of its own. $\ln x$ means $\log_e x$ , where $e$ is the fixed number $2.71828\ldots$ , irrational like $\pi$ and just as ordinary.

Nothing about it is new. It is a base like any other, wearing a different name.

Rule

$\ln x = \log_e x$

$e \approx 2.71828$

$\ln e^n = n$

$e^{\ln x} = x$

How it is used

1. Three notations, one idea

Written	Base	Read as
$\log_a x$	$a$ , stated	the power that turns $a$ into $x$
$\log x$	$10$ , hidden	the power that turns $10$ into $x$
$\ln x$	$e$ , hidden	the power that turns $e$ into $x$

Two of the three hide their base. Which one is hidden is the only difference between the last two rows.

2. The four rules with a = e

\ln 1 = 0

\ln e = 1

\ln e^n = n

e^{\ln x} = x

Substituting $a = e$ is the whole derivation. $e$ is never worked out as a decimal here, and never needs to be.

3. Values you read at sight

Write the argument as a power of $e$ first. The exponent is then the answer.

$x$	as a power of $e$	$\ln x$
$e^4$	$e^4$	$4$
$e$	$e^1$	$1$
$1$	$e^0$	$0$
$\sqrt{e}$	$e^{1/2}$	$\dfrac{1}{2}$
$\dfrac{1}{e}$	$e^{-1}$	$-1$
$\dfrac{1}{e^3}$	$e^{-3}$	$-3$

The domain has not moved either: $x > 0$ , so $\ln 0$ and $\ln(-2)$ still have no value.

4. Why this base at all

$e$ is the base that shows up whenever a quantity grows at a rate proportional to its own size, and it is the base calculus will use when it starts differentiating logarithms. For now that is a promise, not a calculation.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 2.718 · Blue: y = 2.718ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -0.693.

Worked example

Evaluate $\ln e^5 - \ln 1 + e^{\ln 3}$ .

Remember

$\ln$ means $\log_e$ , a logarithm in base $e$ $\ln e^n = n$ , so read the exponent straight off $e^{\ln x} = x$ , the two cancel

Take the first term. Its argument is already a power of $e$ .

\ln e^5 = 5

$\ln 1 = 0$ , because $e^0 = 1$ .

5 - 0 + e^{\ln 3}

In the last term the exponential and the logarithm cancel.

e^{\ln 3} = 3

Add what is left.

5 - 0 + 3 = 8

Explore

The base is set to $2.718$ , which is $e$ to three decimal places. Drag the probe along the green curve and read the pair of coordinates: at $x = e$ the height is $1$ , at $x = 1$ the height is $0$ .

Now move the base slider away from $2.718$ and back. The curve shifts steepness but keeps every feature: through $(1, 0)$ , climbing forever, never reaching the left of the $y$ -axis. $e$ is not a special shape, only a special number.