The Natural Logarithm

Orientation

The antiderivative of 1/x, which the power rule cannot produce

The power rule for integration handles every exponent but one:

\int t^{n} d t = \frac{t^{n + 1}}{n + 1} + C, n \neq - 1 .

At $n = - 1$ the formula divides by zero. Yet nothing is wrong with the function $1 / t$ or with the area beneath it: on $[1, 2]$ that area is a perfectly definite number, close to $0.693147$ . There is simply no power of $t$ whose derivative is $1 / t$ , so the antiderivative cannot be written with the functions available so far.

The response is to stop searching and make a definition:

\ln x = \int_{1}^{x} \frac{d t}{t}, x > 0 .

This can feel like naming a difficulty rather than solving it. It is not, and the reason matters: each $\ln x$ is a specific number, computable to any accuracy by the methods of the integration unit, and every property of the function can now be derived rather than assumed.

What comes out of it. Three things, none of them assumed in advance.

The product law is geometry. Splitting $\int_{1}^{a b}$ at $a$ and rescaling the second piece shows $\ln (a b) = \ln a + \ln b$ . The property that made logarithms worth inventing, turning multiplication into addition, is a fact about this area. Checked: $\ln 6 = 1.791759469228$ and $\ln 2 + \ln 3 = 1.791759469228$ .

The number $e$ is located, not chosen. The area grows steadily as the upper limit moves right, so at exactly one point it equals 1. That point is $e$ , found by bisection to be $2.718281828459$ . Nothing was picked; $e$ was discovered as the answer to "where does the area reach 1?"

The exponential comes last. $\ln$ is strictly increasing, so it has an inverse, and that inverse is $e^{x}$ . This reverses the familiar order, usually $e^{x}$ comes first and $\ln$ is its inverse, but the familiar order must assume what $e$ is and what $e^{x}$ means for irrational $x$ . Here both are supplied.

Where this function already appears. The integration-techniques unit produces $\ln | u |$ constantly from substitutions and partial fractions; the functions unit gives $\ln g (x)$ the domain condition $g (x) > 0$ ; l'Hôpital's rule uses $\ln$ to handle indeterminate powers. All of that rests on the definition given here.

Why this matters

Why define a function by an integral at all

Defining a function as an area looks like a retreat. Why not define $\ln$ as the inverse of $e^{x}$ , the way it is usually first met?

Because that route has a gap in it. To define $\ln$ as the inverse of $e^{x}$ , one must first know what $e^{x}$ means. For integer exponents $e^{3} = e \cdot e \cdot e$ is clear, and rational exponents follow through roots. But $e^{\sqrt{2}}$ is not a product of anything, and the usual account, "take a limit of rational exponents", needs the very machinery of limits and continuity that the logarithm is often used to develop. It also needs $e$ itself to have been produced from somewhere, and the usual source is $lim (1 + 1 / n)^{n}$ , which converges so slowly it gives 2.718280469096 only after a million terms.

The integral route has no such gap. The area under $1 / t$ exists for every $x > 0$ because $1 / t$ is continuous there, and the fundamental theorem does the rest.

What the definition supplies, in order.

Step	Obtained from	Requires no prior notion of
$\ln x$ exists for all $x > 0$	continuity of $1 / t$	exponentials
$\frac{d}{d x} \ln x = 1 / x$	fundamental theorem	limits of rational powers
$\ln (a b) = \ln a + \ln b$	splitting and rescaling the integral	any law of exponents
$e$ with $\ln e = 1$	monotonicity and the intermediate value theorem	$e$ being given in advance
$e^{x}$ for every real $x$	inverse of $\ln$	irrational exponents

Each row uses only what is above it. In particular $e^{\sqrt{2}}$ is now unproblematic: it is the number whose logarithm is $\sqrt{2}$ , and no limit of rational powers is needed to say what it means.

The general pattern. Defining a function by an integral is standard practice whenever the object wanted has no elementary closed form. The error function $\erf (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t$ is defined this way for exactly the reason the series unit gives, $e^{- t^{2}}$ has no elementary antiderivative, and so is the logarithmic integral used in number theory. A definition of this kind is not an admission of defeat; it specifies a number for each input and permits every subsequent property to be proved.

The cost. Values are not obvious by inspection. $\ln 2$ is not a familiar constant the way $\sqrt{2}$ is; it is $0.693147180560$ , obtained by computing an area. And the series $\ln (1 + u) = u - \frac{u^{2}}{2} + \frac{u^{3}}{3} - \dots$ that could supply values converges only for $- 1 < u \leq 1$ , uselessly slowly at the endpoint: 10000 terms of the alternating harmonic series give $0.6930971831$ against $\ln 2 = 0.6931471806$ , still wrong in the fifth decimal.

So the definition is excellent for proving things and poor for computing them, which is why the laws derived from it matter so much, since they reduce awkward logarithms to combinations of a few known values.

Definition

Sign, domain, and what the definition does not assume

The lower limit is 1, and that choice fixes everything. Starting the integral at 1 makes $\ln 1 = 0$ , since an integral over an empty interval vanishes. Any other starting point would shift the whole function by a constant and break the product law, $\ln (a b) = \ln a + \ln b$ at $a = b = 1$ forces $\ln 1 = 2 \ln 1$ , hence $\ln 1 = 0$ . So the lower limit is not a convention but the only choice compatible with the law.

The sign comes from orientation, not from the integrand. For $x > 1$ the integral runs left to right over a positive integrand, so $\ln x > 0$ . For $0 < x < 1$ the limits are reversed, and $\int_{1}^{x} = - \int_{x}^{1}$ , so $\ln x < 0$ . The integrand $1 / t$ is positive throughout; the negative values of $\ln$ come entirely from the direction of travel.

Checked: $\ln 2 = + 0.693147180560$ and $\ln 0.5 = - 0.693147180560$ , equal in size, opposite in sign, which is the quotient law at $\ln (1 / 2) = - \ln 2$ .

Why the domain is $x > 0$ , exactly. The integrand has a singularity at $t = 0$ . For $x > 0$ the path from 1 to $x$ stays within $(0, \infty)$ and never meets it. For $x < 0$ the path would have to cross $t = 0$ , where $1 / t$ is unbounded and the integral does not exist. That is the source of the condition $g (x) > 0$ the functions unit imposes on $\ln g (x)$ . A domain restriction inherited from an integral, not an arbitrary rule.

Note the related but different statement $\int \frac{d u}{u} = \ln | u | + C$ . The absolute value is legitimate because on an interval of negative numbers $\frac{d}{d u} \ln (- u) = \frac{- 1}{- u} = \frac{1}{u}$ as well. But this holds on any interval not containing 0; an integral spanning 0 still fails to exist. Verified: $\int_{- 2}^{- 1} \frac{d t}{t} = - 0.693147180560$ , matching $\ln | - 1 | - \ln | - 2 | = 0 - \ln 2$ .

What is deliberately not assumed. The definition mentions no exponential, no base, and no value of $e$ . That is the point of the ordering: $e$ is defined after $\ln$ , as the solution of $\ln x = 1$ , and $e^{x}$ after that, as the inverse function. Reading the definition as "the logarithm to base $e$ " inverts the logic and smuggles in the thing being constructed.

The strict increase that makes the inverse exist is itself a consequence: $\frac{d}{d x} \ln x = \frac{1}{x} > 0$ for all $x > 0$ . Verified numerically at $x = 0.5, 1, 2, 5, 10$ , where the central difference gives $2.0$ , $1.0$ , $0.5$ , $0.2$ , $0.1$ to ten decimals.

Figure

The logarithm is the area under 1/t

The area from 1 to e under 1/t, which is exactly 1

$\ln x$ is defined as $\int_{1}^{x} \frac{d t}{t}$ , and the shaded strip is that area for $x = e$ , where it equals exactly 1.

Two things the picture does settle. The area starts at $t = 1$ , so $\ln 1 = 0$ : an empty interval has no area. And for $x > 1$ the strip lies to the right of the start and counts positively, so $\ln$ is increasing there: each extra bit of $x$ adds a positive sliver of area, of height $1 / x$ .

The sign below 1 follows from the same definition rather than from the drawing: for $x < 1$ the interval runs backwards, $\int_{1}^{x} = - \int_{x}^{1}$ , so $\ln x$ is negative. That is an orientation convention made visible only if you draw the second interval, which this figure does not.

The product law and the unboundedness of $\ln$ are established in the derivation and theorem blocks that follow. Neither is readable off this strip: a curve approaching the axis may enclose finite or infinite area, and which one happens for $1 / t$ takes an argument, not a glance.

Derivation

Deriving the product and power laws from the integral

The product law. Claim: for $a, b > 0$ , $\ln (a b) = \ln a + \ln b$ .

Start from the definition and split the interval at $a$ :

\ln (a b) = \int_{1}^{a b} \frac{d t}{t} = \int_{1}^{a} \frac{d t}{t} + \int_{a}^{a b} \frac{d t}{t} = \ln a + \int_{a}^{a b} \frac{d t}{t} .

The claim reduces to showing the second piece equals $\ln b$ , that the area from $a$ to $a b$ is the same as the area from 1 to $b$ .

The rescaling. Substitute $t = a u$ , so $d t = a d u$ , and the limits $t = a, a b$ become $u = 1, b$ :

\int_{a}^{a b} \frac{d t}{t} = \int_{1}^{b} \frac{a d u}{a u} = \int_{1}^{b} \frac{d u}{u} = \ln b .

The factor $a$ cancels exactly, appearing once from $d t$ and once in the denominator. Hence

\ln (a b) = \ln a + \ln b . ◼

What made it work. Only $1 / t$ has this property. Stretching the interval by a factor $a$ multiplies its width by $a$ and divides the height by $a$ , so the area is unchanged. For any other power the two factors would not cancel: under $t = a u$ , $t^{n}$ picks up $a^{n + 1}$ , which is 1 only when $n = - 1$ . The product law is a property of the reciprocal, not of areas in general, which is why exactly one power lacks an elementary antiderivative and exactly one gives a function converting products to sums.

Verified: $\ln 6 = 1.791759469228$ against $\ln 2 + \ln 3 = 1.791759469228$ ; $\ln 8 = 2.079441541680$ against $\ln 2 + \ln 4 = 2.079441541680$ ; $\ln 35 = 3.555348061489$ against $\ln 5 + \ln 7 = 3.555348061489$ ; and $\ln 4 = 1.386294361120$ against $\ln 0.5 + \ln 8 = 1.386294361120$ , which covers a case with one argument below 1.

The quotient law follows. Applying the product law to $b \cdot \frac{a}{b} = a$ gives $\ln b + \ln \frac{a}{b} = \ln a$ , so $\ln \frac{a}{b} = \ln a - \ln b$ . With $a = 1$ : $\ln \frac{1}{b} = - \ln b$ , which is why $\ln 0.5 = - \ln 2$ exactly.

The power law, for integers first. Repeated use of the product law gives

\ln (a^{n}) = \underset{n}{\underset{⏟}{\ln a + \dots + \ln a}} = n \ln a

for positive integers $n$ . Verified: $\ln (2^{n})$ against $n \ln 2$ at $n = 1, 2, 4, 8, 10$ , $0.693147180560$ , $1.386294361120$ , $2.772588722240$ , $5.545177444480$ , $6.931471805599$ , matching in every digit shown.

The power law for all real $r$ , which repeated multiplication cannot reach, comes from calculus instead. Fix $r$ and differentiate $\ln (x^{r})$ by the chain rule:

\frac{d}{d x} \ln (x^{r}) = \frac{1}{x^{r}} \cdot r x^{r - 1} = \frac{r}{x} = \frac{d}{d x} (r \ln x) .

Two functions with the same derivative on an interval differ by a constant, and at $x = 1$ both are zero, so the constant is zero and $\ln (x^{r}) = r \ln x$ for every real $r$ . This is the step that makes $\ln (2^{\sqrt{2}}) = \sqrt{2} \ln 2$ meaningful.

Locating $e$ . Since $\ln$ is continuous with $\ln 1 = 0$ , and $\ln (2^{n}) = n \ln 2$ grows without bound, the intermediate value theorem gives exactly one $x$ with $\ln x = 1$ . Bisecting on the condition $\int_{1}^{x} d t / t = 1$ yields

e = 2.718281828459,

agreeing with the known value to all twelve places shown. Strict monotonicity makes it unique, so $e$ is the output of a search, not a choice.

Theorem

The logarithm as a bijection onto the reals

Theorem. $\ln : (0, \infty) \to R$ is strictly increasing, continuous, and onto. Consequently it has an inverse defined on all of $R$ , denoted $\exp$ or $e^{x}$ , satisfying $\ln (e^{x}) = x$ for every real $x$ and $e^{\ln x} = x$ for every $x > 0$ .

Strictly increasing. By the fundamental theorem, $\frac{d}{d x} \ln x = \frac{1}{x}$ , which is positive for every $x > 0$ . A function with positive derivative on an interval is strictly increasing there.

Verified numerically: central differences give $2.0$ , $1.0$ , $0.5$ , $0.2$ , $0.1$ at $x = 0.5, 1, 2, 5, 10$ , matching $1 / x$ to within $2 \times 10^{- 10}$ .

Continuous. Differentiability implies continuity.

Unbounded above. The power law gives $\ln (2^{n}) = n \ln 2$ , and $\ln 2 = 0.693147180560 > 0$ , so choosing $n$ large makes $\ln (2^{n})$ exceed any bound.

Unbounded below. $\ln \frac{1}{2^{n}} = - n \ln 2 \to - \infty$ by the quotient law.

Onto. Being continuous and taking arbitrarily large and arbitrarily negative values, $\ln$ attains every real value in between, by the intermediate value theorem. Strict increase makes the attainment unique, so the function is a bijection and the inverse exists.

Verified: $\ln (e^{x}) = x$ and $e^{\ln x} = x$ at $x = 0.5, 1, 2, 5$ , exact to twelve places in every case.

Corollary (the definition of $e$ ). Since 1 is a real number, exactly one $x$ satisfies $\ln x = 1$ . That number is $e = 2.718281828459$ , recovered here by bisecting on the area condition rather than by quoting it.

Corollary (all logarithms are multiples of this one). For $b > 0$ , $b \neq 1$ , define $\log_{b} x$ as the solution $y$ of $b^{y} = x$ . Taking $\ln$ of both sides and using the power law, $y \ln b = \ln x$ , so

\log_{b} x = \frac{\ln x}{\ln b} .

Verified: $\log_{10} 1000 = 3$ , $\log_{2} 1024 = 10$ , $\log_{3} 81 = 4$ , each matching $\ln x / \ln b$ to twelve places. So there is only one logarithm up to a constant factor, and the "natural" one is the choice making the derivative exactly $1 / x$ rather than $\frac{1}{x \ln b}$ .

Corollary (exponentials for irrational powers). For $a > 0$ and any real $r$ , define $a^{r} = e^{r \ln a}$ . This is consistent with repeated multiplication when $r$ is an integer and supplies a meaning when $r$ is irrational, filling the gap the motivation block identifies in the usual order of development.

Growth, and a caution. $\ln x \to \infty$ , but slowly:

$x$	$\ln x$
$10$	$2.302585$
$100$	$4.605170$
$10^{6}$	$13.815511$
$10^{12}$	$27.631021$
$10^{100}$	$230.258509$

Each hundredfold increase in $x$ adds only about $4.6$ . For $\ln x$ to exceed 100 requires $x > e^{100} \approx 2.688 \times 10^{43}$ . So unboundedness and slow growth are compatible. The same lesson the harmonic series teaches in the sequences unit, and for a closely related reason, since $H_{n}$ tracks $\ln n$ .

Procedure

Working with logarithms

To differentiate an expression containing $\ln$ .

Step 1 — Simplify with the laws first. Converting $\ln \frac{x^{2} \sqrt{x + 1}}{x - 3}$ into $2 \ln x + \frac{1}{2} \ln (x + 1) - \ln (x - 3)$ turns one messy chain-rule problem into three easy ones. Do this before differentiating, not after.

Step 2 — Apply the chain rule. For a composite,

\frac{d}{d x} \ln u (x) = \frac{u^{'} (x)}{u (x)} .

The derivative of a logarithm is the derivative of the inside over the inside.

Step 3 — Use logarithmic differentiation where it helps. For a product or quotient of many factors, or for a variable base with a variable exponent such as $x^{x}$ , take $\ln$ of both sides, differentiate implicitly, then multiply back by $y$ . This is the only elementary route to $\frac{d}{d x} x^{x} = x^{x} (1 + \ln x)$ .

To integrate.

Step 4 — Recognise the pattern $\frac{u^{'}}{u}$ . Whenever the numerator is the derivative of the denominator,

\int \frac{u^{'}}{u} d x = \ln | u | + C .

Keep the absolute value: it is what makes the result valid on intervals where $u < 0$ . Check the interval does not contain a zero of $u$ ; if it does, the integral does not exist.

To evaluate or estimate a logarithm.

Step 5 — Reduce to known values with the laws. $\ln 12 = \ln (4 \cdot 3) = 2 \ln 2 + \ln 3$ . Working from $\ln 2 = 0.693147180560$ and $\ln 3 = 1.098612288668$ gives $2.484906649788$ , and $\ln 12$ is indeed.

Step 6 — Convert bases when needed with $\log_{b} x = \frac{\ln x}{\ln b}$ .

To solve an equation containing logarithms.

Step 7 — Record the domain before doing any algebra. Every $\ln g (x)$ in the original equation requires $g (x) > 0$ , and these conditions are the equation's domain.

Step 8 — Combine into a single logarithm, then exponentiate. Use the laws to reach the form $\ln A = \ln B$ or $\ln A = c$ , then apply $e^{(\cdot)}$ to both sides.

Step 9 — Check every candidate against the domain recorded in step 7. Combining logarithms enlarges the domain, so the algebra can produce roots the original equation excludes. Without it the answer is wrong.

Where it goes wrong.

Splitting $\ln (a + b)$ . There is no such law. The laws convert products to sums, never sums to anything.
Dropping the absolute value in $\int \frac{d u}{u}$ , which silently restricts the result to positive $u$ .
Skipping the domain check after combining logarithms. The error the functions unit illustrates with $\ln x + \ln (x - 3) = \ln 10$ , whose algebra yields $x = 5$ and $x = - 2$ though only $x = 5$ lies in the domain $x > 3$ .
Treating $\ln x$ as bounded because it grows slowly. It exceeds every bound eventually, just very late: $\ln x > 100$ needs $x > 2.688 \times 10^{43}$ .
Using the series $\ln (1 + u) = u - \frac{u^{2}}{2} + \dots$ outside $- 1 < u \leq 1$ , where it diverges, or at $u = 1$ for computation, where 10000 terms still give only $0.6930971831$ against $0.6931471806$ .
Applying a law with a non-positive argument. $\ln (a b) = \ln a + \ln b$ needs $a > 0$ and $b > 0$ separately; with $a = b = - 2$ the left side is $\ln 4$ and the right side is undefined.

Worked example

Solving a logarithmic equation with a domain check

Problem. Solve $\ln x + \ln (x - 3) = \ln 10$ .

Step 7 — Record the domain first. The equation contains $\ln x$ and $\ln (x - 3)$ , so it requires

x > 0 and x - 3 > 0 .

The second is stricter, so the domain is $x > 3$ . Writing this down before any algebra is what makes the final check possible. Once the logarithms are combined, the original restrictions are no longer visible.

Step 8 — Combine and exponentiate. By the product law, valid here because both arguments are positive on the domain,

\ln (x (x - 3)) = \ln 10 .

Since $\ln$ is injective, being strictly increasing by the theorem block, equal logarithms force equal arguments:

x (x - 3) = 10 ⟹ x^{2} - 3 x - 10 = 0 ⟹ (x - 5) (x + 2) = 0,

giving the candidates $x = 5$ and $x = - 2$ .

Step 9 — Check both against the domain.

Candidate	In $x > 3$ ?	Verdict
$x = 5$	yes	valid
$x = - 2$	no	rejected

At $x = - 2$ the original equation reads $\ln (- 2) + \ln (- 5)$ , and neither term exists. The root is an artefact of the algebra, not a solution.

Why the algebra produced it. The step from $\ln x + \ln (x - 3)$ to $\ln (x (x - 3))$ is valid only where both original logarithms are defined, that is on $x > 3$ . But the combined form $\ln (x (x - 3))$ is defined wherever the product is positive, which includes $x < 0$ , since two negatives multiply to a positive. The manipulation quietly enlarged the domain, and the extra root came from the enlargement rather than from the equation.

This is the same phenomenon the functions unit describes, and the reason step 7 exists.

Verify the surviving root. At $x = 5$ :

\ln 5 + \ln 2 = 1.609437912434 + 0.693147180560 = 2.302585092994,

and $\ln 10 = 2.302585092994$ . Both sides agree to twelve decimal places, and $5 (5 - 3) = 10$ confirms the product directly.

x = 5

---

A second part: differentiate $y = \ln \frac{x^{2} \sqrt{x + 1}}{x - 3}$ for $x > 3$ .

Step 1 — Simplify with the laws before differentiating.

y = 2 \ln x + \frac{1}{2} \ln (x + 1) - \ln (x - 3) .

The quotient became a subtraction, the product a sum, and the square root a coefficient of $\frac{1}{2}$ , all by the laws derived from the area.

Step 2 — Differentiate term by term, each by the chain rule $\frac{d}{d x} \ln u = \frac{u^{'}}{u}$ :

y^{'} = \frac{2}{x} + \frac{1}{2 (x + 1)} - \frac{1}{x - 3} .

Differentiating the original form directly would require the quotient rule, the product rule and the chain rule together, on a fraction containing a square root. Simplifying first replaced all of that with three elementary derivatives, which is why step 1 of the procedure comes first.

Check at $x = 4$ . The simplified expression gives $\frac{2}{4} + \frac{1}{10} - \frac{1}{1} = 0.5 + 0.1 - 1 = - 0.4$ . Differentiating the original numerically at $x = 4$ gives $- 0.400000$ , agreeing.

Example

Areas, laws and a series, computed

1. The definition, evaluated numerically. Computing $\int_{1}^{x} \frac{d t}{t}$ by Simpson's rule and comparing with the known logarithm:

$x$	Area $\int_{1}^{x} d t / t$	$\ln x$	Difference
$1$	$0.000000000000$	$0.000000000000$	$0$
$2$	$0.693147180560$	$0.693147180560$	$2.2 \times 10^{- 16}$
$e$	$1.000000000000$	$1.000000000000$	$1.1 \times 10^{- 16}$
$3$	$1.098612288668$	$1.098612288668$	$5.1 \times 10^{- 15}$
$4$	$1.386294361120$	$1.386294361120$	$5.8 \times 10^{- 15}$
$8$	$2.079441541680$	$2.079441541680$	$1.8 \times 10^{- 15}$
$0.5$	$- 0.693147180560$	$- 0.693147180560$	$2.2 \times 10^{- 16}$

The definition is not a formal gesture: it computes. Note the row at $x = 0.5$ , where the area is negative because the limits run backwards, and the row at $x = e$ , where the area is exactly 1, that row is the definition of $e$ .

2. The laws, checked. $\ln 4 = 1.386294361120$ and $\ln 2 + \ln 2 = 1.386294361120$ ; $\ln 35 = 3.555348061489$ and $\ln 5 + \ln 7 = 3.555348061489$ ; $\ln 4 = 1.386294361120$ and $\ln 0.5 + \ln 8 = 1.386294361120$ , this last mixing a negative logarithm with a positive one.

The power law across several exponents:

$n$	$\ln (2^{n})$	$n \ln 2$
$1$	$0.693147180560$	$0.693147180560$
$2$	$1.386294361120$	$1.386294361120$
$4$	$2.772588722240$	$2.772588722240$
$8$	$5.545177444480$	$5.545177444480$
$10$	$6.931471805599$	$6.931471805599$

This is what "turns multiplication into addition" means concretely: computing $\ln (2^{10})$ needs one multiplication rather than ten.

3. Change of base. $\log_{10} 1000 = 3$ , $\log_{2} 1024 = 10$ , $\log_{3} 81 = 4$ . Each equal to $\ln x / \ln b$ to twelve decimals. Every logarithm is this one, rescaled.

4. The series, and where it fails. For $| u | < 1$ ,

\ln (1 + u) = u - \frac{u^{2}}{2} + \frac{u^{3}}{3} - \frac{u^{4}}{4} + \dots

At $u = 0.5$ , converging to $\ln 1.5 = 0.405465108108$ :

Terms	Value	Error
4	$0.401041666667$	$4.42 \times 10^{- 3}$
8	$0.405315290179$	$1.50 \times 10^{- 4}$
16	$0.405464803171$	$3.05 \times 10^{- 7}$
32	$0.405465108106$	$2.38 \times 10^{- 12}$

At $u = 0.1$ , converging to $\ln 1.1 = 0.095310179804$ , just 8 terms give $0.095310179702$ , error $1.0 \times 10^{- 10}$ , and 16 terms reach the limit of double precision. Smaller $u$ converges dramatically faster.

At the boundary $u = 1$ it becomes the alternating harmonic series, converging to $\ln 2$ so slowly as to be useless:

Terms	Value	Error
$10$	$0.6456349206$	$4.75 \times 10^{- 2}$
$100$	$0.6881721793$	$4.98 \times 10^{- 3}$
$1000$	$0.6926474306$	$5.00 \times 10^{- 4}$
$10000$	$0.6930971831$	$5.00 \times 10^{- 5}$

Ten thousand terms and still wrong in the fifth decimal. Each tenfold increase in work buys one digit. Compare 32 terms at $u = 0.5$ giving twelve correct digits. The contrast is the interval of convergence at work: the series converges for $- 1 < u \leq 1$ , but at the endpoint only barely, which is why practical computation reduces the argument with the laws first rather than using the series directly.

5. Slow growth. $\ln 10 = 2.302585$ , $\ln 10^{6} = 13.815511$ , $\ln 10^{12} = 27.631021$ , $\ln 10^{100} = 230.258509$ . Each factor of $10^{6}$ adds about $13.8$ . The function is unbounded, yet reaching 100 requires $x > 2.688 \times 10^{43}$ .

Warning

There is no law for the logarithm of a sum

The laws convert products into sums. Reading them backwards, as though $\ln$ distributed over addition, produces the most common error in the subject:

\ln (a + b) \neq \ln a + \ln b .

A single counterexample settles it. Take $a = b = 2$ :

\ln (2 + 2) = \ln 4 = 1.386294361120,

while

\ln 2 + \ln 2 = 0.693147180560 + 0.693147180560 = 1.386294361120 .

These happen to agree, because $2 + 2 = 2 \times 2$ , a coincidence peculiar to 2. Take $a = 1$ , $b = 3$ instead:

\ln (1 + 3) = \ln 4 = 1.386294361120, \ln 1 + \ln 3 = 0 + 1.098612288668 = 1.098612288668 .

Different. The correct statement about $\ln 4$ is $\ln (2 \times 2) = \ln 2 + \ln 2$ , using the product rather than the sum.

Why the error is tempting. $\ln$ is written like a function applied to an expression, and many operations do distribute: $\sqrt{}$ does not, but $\frac{d}{d x}$ does, and $2 (a + b) = 2 a + 2 b$ does. The laws here come from a specific fact about the area under $1 / t$ , that rescaling the interval leaves the area unchanged, and that fact concerns multiplication only. Nothing in the derivation touches addition, so no law about sums can emerge from it.

The right move when facing $\ln (a + b)$ . There is usually nothing to do. If the sum factors, factor it: $\ln (x^{2} + x) = \ln (x (x + 1)) = \ln x + \ln (x + 1)$ for $x > 0$ , which works because the product is what the law applies to.

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Two further traps in the same family.

The laws require each argument positive, not merely the product. With $a = b = - 2$ , the left side $\ln (a b) = \ln 4$ exists while the right side $\ln (- 2) + \ln (- 2)$ does not. So the equation $\ln (a b) = \ln a + \ln b$ can fail by the right side being undefined even when the left is fine, and the law must be quoted with its condition $a, b > 0$ .

Combining logarithms enlarges the domain. This is the same asymmetry seen from the other direction, and it is the practical reason equations need a domain check. The step

\ln x + \ln (x - 3) ⟶ \ln (x (x - 3))

is valid on $x > 3$ , where both original terms exist. But the right side is defined on $x < 0$ as well, where the product of two negatives is positive. So the manipulation is not reversible, and solutions found afterwards may lie outside the original domain: $\ln x + \ln (x - 3) = \ln 10$ yields $x = 5$ and $x = - 2$ , and only $x = 5$ is a solution.

The shape of the rule. Each law has a condition, and the condition is always about the arguments being positive, because that is where $\ln$ is defined at all. When a manipulation appears to have produced something from nothing, whether an extra root or a defined expression from an undefined one, the enlargement of the domain is where to look.

Application

Where turning products into sums is the whole point

Every use below exploits the same property: $\ln$ converts multiplication into addition, and repeated multiplication into a coefficient.

Exponential decay and half-lives. A quantity obeying $y^{'} = k y$ has $y = y_{0} e^{k t}$ , as the first-order differential equations unit shows. Asking when it halves means solving $e^{k t} = \frac{1}{2}$ , and the logarithm is what inverts the exponential:

t = \frac{\ln (1 / 2)}{k} = \frac{- \ln 2}{k} = \frac{- 0.693147180560}{k} .

Every half-life calculation carries that constant. For carbon-14 with $k = - 1.21 \times 10^{- 4}$ per year, the half-life is $0.693147 / 1.21 \times 10^{- 4} \approx 5728$ years.

Orders of magnitude. Because $\ln$ turns a factor into an increment, quantities spanning many powers of ten become manageable on a logarithmic scale: pH, decibels, stellar magnitudes, the Richter scale. An earthquake ten times larger is one unit higher, since $\ln (10 x) = \ln x + \ln 10$ and the increment is the same constant $2.302585$ wherever it is applied.

This is the original motivation for logarithms, from the era before mechanical computation: multiplying two seven-digit numbers by hand is laborious, adding their logarithms is not. The slide rule is that idea in wood.

Likelihood in statistics. The probability of independent observations is a product of many small numbers, which underflows to zero in floating-point arithmetic once there are a few hundred factors. Taking $\ln$ converts it to a sum:

\ln \prod_{i} p_{i} = \sum_{i} \ln p_{i},

which is numerically stable and, since $\ln$ is strictly increasing, has its maximum at exactly the same parameter values. Maximum likelihood is almost always computed as maximum log-likelihood for these two reasons together. One about arithmetic, one about the monotonicity established in the theorem block.

Algorithmic complexity. An algorithm halving its input each step finishes in about $\log_{2} n$ steps, because the question "how many halvings reach 1?" is exactly what a logarithm answers. Since all logarithms differ by a constant factor, $\log_{2} n = \ln n / \ln 2$ , complexity classes do not distinguish bases, and $O (\log n)$ is written without one.

The slow growth is what makes such algorithms valuable: $\ln 10^{12} = 27.63$ , so binary search over a trillion items needs about forty comparisons.

Information and entropy. The information content of an event of probability $p$ is $- \ln p$ , chosen precisely so that independent events add: two events of probability $p$ and $q$ jointly have probability $p q$ and information $- \ln (p q) = - \ln p - \ln q$ . The additivity is not a modelling convenience but a consequence of the product law derived from the area.

Integration. Inside mathematics, $\ln$ is the answer to the single gap in the power rule, so it appears whenever an integrand has the form $u^{'} / u$ . The integration-techniques unit produces $\ln | u |$ from substitutions and partial fractions repeatedly, and uses $\ln x$ as the factor to differentiate under integration by parts, all resting on the definition given here.

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The common thread. In each case something multiplicative is awkward, too large to compute, too small to represent, spanning too many scales, or lacking an antiderivative, and $\ln$ converts it into something additive. That conversion is the theorem derived in this unit from the invariance of the area under $1 / t$ when the interval is rescaled.

The Natural Logarithm

The antiderivative of 1/x, which the power rule cannot produce

Why define a function by an integral at all

Sign, domain, and what the definition does not assume

The logarithm is the area under 1/t

Deriving the product and power laws from the integral

The logarithm as a bijection onto the reals

Working with logarithms

Solving a logarithmic equation with a domain check

Areas, laws and a series, computed

There is no law for the logarithm of a sum

Where turning products into sums is the whole point

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