What Is The Natural Logarithm?
The natural logarithm of a positive number $x$ is the power to which $e$ must be raised to give $x$. In symbols, $\ln x = \log_e x$, where $e \approx 2.71828$ is Euler's number. So $\ln x = y$ means exactly $e^{y} = x$.
That makes the natural logarithm the inverse of the exponential function $e^{x}$. Feeding a number into $e^{x}$ and then into $\ln$ returns the original number, and the two undo each other:
$$\ln(e^{x}) = x, \qquad e^{\ln x} = x \quad (x > 0)$$
Three values anchor the whole function:
$\ln 1 = 0$, because $e^{0} = 1$.
$\ln e = 1$, because $e^{1} = e$.
$\ln x$ is only defined for $x > 0$; you cannot take the logarithm of zero or a negative number in the real numbers.
The algebra of $\ln$ as "the log with base $e$" is covered on the companion page for logarithmic functions. This page takes the calculus view: where $\ln$ comes from as an area, how it differentiates and integrates, and why it earns the word "natural."
How Is The Natural Logarithm Defined As An Integral?
Calculus gives the natural logarithm a definition that needs no mention of exponents at all. For $x > 0$:
$$\ln x = \int_{1}^{x} \frac{1}{t},dt$$
Read geometrically, $\ln x$ is the signed area under the curve $y = \tfrac{1}{t}$ between $t = 1$ and $t = x$. This one picture explains the anchor values instantly.
Why $\ln 1 = 0$. The area from $1$ to $1$ has zero width, so $\ln 1 = \int_{1}^{1}\frac{1}{t},dt = 0$.
Why $\ln e = 1$. The number $e$ is defined as the point where the running area first reaches exactly $1$: $\int_{1}^{e}\frac{1}{t},dt = 1$. That is what makes $e \approx 2.71828$ special rather than arbitrary.
Why $\ln x < 0$ for $0 < x < 1$. Integrating "backwards" from $1$ down to a smaller number flips the sign, so the logarithm of a number below $1$ is negative.
What Is The Derivative Of The Natural Logarithm?
The integral definition pays off immediately. By the Fundamental Theorem of Calculus, differentiating an "area so far" function returns the curve you were measuring under. Since $\ln x$ is the area under $\tfrac{1}{t}$, its derivative is $\tfrac{1}{x}$:
$$\frac{d}{dx}[\ln x] = \frac{1}{x}, \qquad x > 0$$
This is the simplest derivative of any standard function, and it is the geometric heart of the topic: the slope of the tangent to the $\ln$ curve at any point $x$ is exactly $\tfrac{1}{x}$. Near $x=1$ the graph climbs with slope $1$; far to the right it flattens, because $\tfrac{1}{x}$ shrinks toward $0$.
For negative inputs the absolute-value version holds for every $x \neq 0$:
$$\frac{d}{dx}\big[\ln|x|\big] = \frac{1}{x}$$
The check is quick. For $x > 0$, $\ln|x| = \ln x$, whose derivative is $\tfrac{1}{x}$. For $x < 0$, $\ln|x| = \ln(-x)$, and the chain rule gives $\frac{1}{-x}\cdot(-1) = \tfrac{1}{x}$. Both cases land on $\tfrac{1}{x}$.
Most real problems wrap the logarithm around another function. The chain rule then gives the single most-used differentiation formula in this topic:
$$\frac{d}{dx}\big[\ln(g(x))\big] = \frac{g'(x)}{g(x)}$$
In words: the derivative of the log of something is the derivative of the inside, divided by the inside. Forgetting that inner derivative $g'(x)$ is the classic slip, and it has its own entry in the mistakes section below.
How Do You Integrate 1/x With The Natural Logarithm?
The power rule $\int x^{n},dx = \frac{x^{n+1}}{n+1} + C$ breaks at $n = -1$, because it would divide by zero. The natural logarithm is exactly the function that fills that hole:
$$\int \frac{1}{x},dx = \ln|x| + C$$
The absolute value matters. The integrand $\tfrac{1}{x}$ is defined for negative $x$ too, so the antiderivative must be as well, and $\ln|x|$ covers both sides of zero. Differentiating $\ln|x| + C$ returns $\tfrac{1}{x}$, which confirms the formula.
Wrapped around an inner function, this becomes the pattern that solves a huge share of integrals, any fraction whose numerator is the derivative of its denominator:
$$\int \frac{g'(x)}{g(x)},dx = \ln|g(x)| + C$$
What Are The Log Laws Of The Natural Logarithm?
Because $\ln$ is a logarithm, it turns products into sums, quotients into differences, and powers into multiples. For $a, b > 0$ and any real $r$:
Table: The natural-logarithm laws, with the everyday meaning of each.
Law | Statement | What it does |
|---|---|---|
Product | $\ln(ab) = \ln a + \ln b$ | Multiplication becomes addition |
Quotient | $\ln!\left(\dfrac{a}{b}\right) = \ln a - \ln b$ | Division becomes subtraction |
Power | $\ln(a^{r}) = r,\ln a$ | An exponent comes out front as a factor |
Identity | $\ln 1 = 0,\quad \ln e = 1$ | The two anchor values |
These are not accidents of the base; the integral definition proves them. Take the product law. Define $f(x) = \ln(ax)$ for a fixed constant $a > 0$. By the chain rule, $f'(x) = \frac{1}{ax}\cdot a = \frac{1}{x}$, which is the same derivative as $\ln x$.
Two functions with the same derivative differ only by a constant, so $\ln(ax) = \ln x + C$. Setting $x = 1$ gives $\ln a = 0 + C$, so $C = \ln a$. Therefore $\ln(ax) = \ln x + \ln a$, and putting $x = b$ yields $\ln(ab) = \ln a + \ln b$; the other laws follow the same way. For the full algebraic treatment, see logarithm rules and properties of logarithms.
How Do You Change The Base To The Natural Logarithm?
Any logarithm can be rewritten in terms of the natural logarithm using the change-of-base formula. For a base $b > 0$, $b \neq 1$:
$$\log_{b} x = \frac{\ln x}{\ln b}$$
This is why a calculator with only $\ln$ and $\log_{10}$ keys can still find, say, $\log_{2} 10$:
$$\log_{2} 10 = \frac{\ln 10}{\ln 2} = \frac{2.302585}{0.693147} = 3.3219$$
Change of base also explains the derivative of a general logarithm. Writing $\log_{b} x = \frac{\ln x}{\ln b}$ and differentiating (with $\ln b$ a constant) gives:
$$\frac{d}{dx}\big[\log_{b} x\big] = \frac{1}{x,\ln b}$$
The extra $\ln b$ in the denominator is the "tax" for using any base other than $e$. Base $e$ is the only one where that factor equals $1$ and the derivative is simply $\tfrac{1}{x}$, which is a first hint at why it is called natural. Rewriting between exponential and log form is covered on the exponential form page.
How Do You Work With The Natural Logarithm? (Worked Examples)
Each example below is checked by differentiating the answer back to the starting expression, or by re-deriving the value.
Example 1: Differentiate $y = \ln(3x^{2} + 1)$.
Use the chain rule $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$ with $g(x) = 3x^{2} + 1$, so $g'(x) = 6x$:
$$\frac{dy}{dx} = \frac{6x}{3x^{2} + 1}$$
Check: the inside derivative is $6x$ and the inside is $3x^{2}+1$, matching the formula.
Final answer: $\dfrac{dy}{dx} = \dfrac{6x}{3x^{2} + 1}$.
Example 2: Differentiate $y = \ln(\sin x)$.
Here $g(x) = \sin x$ and $g'(x) = \cos x$:
$$\frac{dy}{dx} = \frac{\cos x}{\sin x} = \cot x$$
Final answer: $\dfrac{dy}{dx} = \cot x$. (This is one of the ways the natural logarithm connects to trigonometry.)
Example 3: Evaluate $\displaystyle\int \frac{2x}{x^{2} + 1},dx$.
The numerator $2x$ is exactly the derivative of the denominator $x^{2} + 1$, so this fits $\int \frac{g'(x)}{g(x)},dx = \ln|g(x)| + C$. Let $u = x^{2} + 1$, so $du = 2x,dx$:
$$\int \frac{2x}{x^{2} + 1},dx = \int \frac{1}{u},du = \ln|u| + C = \ln(x^{2} + 1) + C$$
Because $x^{2} + 1 > 0$ for every real $x$, the absolute-value bars are optional here.
Check: $\dfrac{d}{dx}\big[\ln(x^{2}+1) + C\big] = \dfrac{2x}{x^{2}+1}$, the original integrand.
Final answer: $\ln(x^{2} + 1) + C$.
Example 4: Evaluate the definite integral $\displaystyle\int_{1}^{e} \frac{1}{x},dx$.
$$\int_{1}^{e} \frac{1}{x},dx = \Big[\ln|x|\Big]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1$$
Final answer: $1$. This is the definition of $e$ read as an area: the region under $\tfrac{1}{x}$ from $1$ to $e$ has area exactly $1$.
Example 5: Find $x$ if $\ln x = 2$, to four decimal places.
Rewrite in exponential form: $\ln x = 2$ means $x = e^{2}$.
$$x = e^{2} = 2.71828^{2} = 7.3891$$
Check: $\ln 7.3891 \approx 2$. Final answer: $x = 7.3891$.
A reference table collects the results used above.
Table: Core derivative and integral forms for the natural logarithm.
Expression | Result | Note |
|---|---|---|
$\dfrac{d}{dx}[\ln x]$ | $\dfrac{1}{x}$ | $x > 0$ |
$\dfrac{d}{dx}[\ln | x | ]$ |
$\dfrac{d}{dx}[\ln(g(x))]$ | $\dfrac{g'(x)}{g(x)}$ | Chain rule |
$\displaystyle\int \dfrac{1}{x},dx$ | $\ln | x |
$\displaystyle\int \dfrac{g'(x)}{g(x)},dx$ | $\ln | g(x) |
For the wider families these belong to, see the derivative and integration formula pages.
Why Are e And The Natural Logarithm Called "Natural"?
The name is not marketing. Base $e$ is the one choice that removes every stray constant from calculus, and the reasons stack up.
The exponential is its own derivative. Only $y = e^{x}$ satisfies $\frac{dy}{dx} = e^{x}$; its rate of change equals its current value at every point. For any other base, $\frac{d}{dx}[b^{x}] = b^{x}\ln b$ carries an extra $\ln b$ factor.
The derivative of $\ln$ is clean. As shown above, $\frac{d}{dx}[\log_{b} x] = \frac{1}{x\ln b}$, and only $b = e$ makes that factor $1$, leaving the tidy $\tfrac{1}{x}$.
It is the mathematics of continuous growth. Anything that grows or decays at a rate proportional to its current size (money compounding, a population, a cooling body, a decaying isotope) is governed by $e^{x}$, and reading the time or rate back out of it requires $\ln$. The constant itself is the limit of continuous compounding, $e = \displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}$.
So $\ln$ is "natural" in the precise sense that it is the logarithm calculus would invent on its own, with no arbitrary base bolted on. It is the inverse of the one exponential whose growth rate is its own size. The link to exponential and inverse behaviour is developed on the inverse functions page.
Who Discovered The Natural Logarithm?
The natural logarithm arrived twice: first as a mysterious area under a curve, and only later as "the log with base $e$." The area came first.
Two more figures completed the picture:
John Napier (1550–1617, Scotland) published the first logarithm tables in 1614, inventing logarithms as a tool to turn multiplication into addition for astronomers and navigators.
Leonhard Euler (1707–1783, Switzerland) named the constant $e$, showed that $e = \lim_{n\to\infty}(1+\tfrac{1}{n})^{n}$, and made $\ln$ the inverse of $e^{x}$ the standard view. The base of the natural logarithm carries his initial to this day.
Where Is The Natural Logarithm Used In The Real World?
Any process with a rate proportional to its own size is governed by $e$ and read back with $\ln$.
Radioactive decay and carbon dating. A sample decays as $N = N_{0}e^{-\lambda t}$, and the half-life is $t_{1/2} = \dfrac{\ln 2}{\lambda}$. The natural logarithm is what converts a decay rate into an age.
Continuous growth and finance. Continuously compounded money grows as $A = Pe^{rt}$; the time for an investment to double is $\dfrac{\ln 2}{r}$, the same $\ln 2 \approx 0.6931$ that appears in half-life.
Information theory. Entropy measured in nats uses the natural logarithm, $H = -\sum p,\ln p$, quantifying uncertainty in a signal or a message.
Chemistry and seismology. The pH scale and the Richter magnitude scale are logarithmic, compressing enormous ranges into readable numbers; the same logarithmic compression, in base $e$, drives reaction-rate and cooling models.
Machine learning. Log-likelihood and cross-entropy loss functions are built on $\ln$, because turning products of probabilities into sums of logarithms keeps the numbers stable and the calculus simple.
One function connects an isotope's age, a bank balance, and a message's information content, because all three change at a rate set by their own size.
What Are The Most Common Mistakes With The Natural Logarithm?
These four errors account for most lost marks on the natural logarithm, verified against Johns Hopkins and UConn logarithm notes, Britannica's natural-log entry, and the recurring r/learnmath and r/calculus questions behind them.
Splitting the log of a sum.
Where it slips in:
A student rewrites $\ln(a + b)$ as $\ln a + \ln b$, borrowing the product law and applying it to a sum.
Don't do this:
There is no rule that simplifies $\ln(a + b)$. The product law only turns a product into a sum: $\ln(ab) = \ln a + \ln b$, never $\ln(a+b)$.
The correct way:
Leave $\ln(a + b)$ as it is. Only split a logarithm when the inside is genuinely a product, quotient, or power.
Dropping the $+C$ or the absolute value in $\int \frac{1}{x},dx$.
Where it slips in:
A student writes $\int \frac{1}{x},dx = \ln x$, missing both the constant of integration and the absolute-value bars.
Don't do this:
Do not write $\ln x$ alone. Without $+C$ the antiderivative is incomplete, and without the bars it is undefined for negative $x$.
The correct way:
Always write $\int \frac{1}{x},dx = \ln|x| + C$. The bars keep it valid on both sides of zero, and the $+C$ belongs on every indefinite integral.
Forgetting the inner derivative in the chain rule.
Where it slips in:
Differentiating $\ln(3x^{2}+1)$, a student writes $\frac{1}{3x^{2}+1}$ and stops, leaving out the derivative of the inside.
Don't do this:
Do not treat $\frac{d}{dx}[\ln(g(x))]$ as just $\frac{1}{g(x)}$. That is only half of the chain rule.
The correct way:
Multiply by the inner derivative: $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$. For $\ln(3x^{2}+1)$ the answer is $\frac{6x}{3x^{2}+1}$.
Confusing $\ln(x^{2})$ with $(\ln x)^{2}$.
Where it slips in:
A student treats "$\ln x$ squared" as if the two expressions were interchangeable when applying the power law.
Don't do this:
Do not write $(\ln x)^{2} = 2\ln x$. The power law applies to the exponent inside the logarithm, not to the whole logarithm raised to a power.
The correct way:
Keep them separate: $\ln(x^{2}) = 2\ln x$ (exponent inside comes out front), while $(\ln x)^{2}$ is the logarithm multiplied by itself and does not simplify.
Practice Problems On The Natural Logarithm
Answers follow each problem; every derivative and integral checks against the rules above.
Differentiate $y = \ln(5x)$.
(Answer: $\ln(5x) = \ln 5 + \ln x$, so $\frac{dy}{dx} = \frac{1}{x}$; equivalently $\frac{5}{5x} = \frac{1}{x}$.)Differentiate $y = \ln(x^{3} + 2x)$.
(Answer: $\frac{dy}{dx} = \frac{3x^{2} + 2}{x^{3} + 2x}$.)Differentiate $y = x\ln x$.
(Answer: product rule gives $\frac{dy}{dx} = \ln x + x\cdot\frac{1}{x} = \ln x + 1$.)Evaluate $\displaystyle\int \frac{3}{x},dx$.
(Answer: $3\ln|x| + C$.)Evaluate $\displaystyle\int_{1}^{e^{2}} \frac{1}{x},dx$.
(Answer: $\ln(e^{2}) - \ln 1 = 2 - 0 = 2$.)Use change of base to find $\log_{2} 10$ to four decimal places.
(Answer: $\frac{\ln 10}{\ln 2} = \frac{2.302585}{0.693147} = 3.3219$.)
Where Should You Go Next After The Natural Logarithm?
The natural logarithm sits at the junction of algebra and calculus, and several doors open from here.
Logarithmic functions. The algebra side: graphs, domains, and how $\ln$ fits the wider family of logarithmic functions.
Logarithm rules. Drill the product, quotient, and power laws that this page proved from the integral.
Calculus. Step back to the big picture of derivatives and integrals, where $\ln$ and $\frac{1}{x}$ first meet.
If your child is building these foundations, a live Bhanzu trainer teaches the natural logarithm from the "why" (the area under $\tfrac{1}{t}$ and the growth it measures) in the Bhanzu math classes.
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