Of all the possible bases for logarithms (2, 10, 7.3, ...), one stands out: the number $e \approx 2.71828$. The logarithm with base $e$ is so important that it gets its own name and notation:
$$\ln x = \log_e x$$
But why is base $e$ special? The full answer will become clear when you study derivatives in Section 6.4, but here's a preview: the derivative of $\ln x$ is simply $\frac{1}{x}$, with no messy constants. This makes $\ln x$ the "natural" choice for calculus.
For now: Think of $\ln$ as "the logarithm that calculus likes best."
Self-check: Can you answer these questions? If not, review the linked prerequisites first.
| Question | If you struggle... |
|---|---|
| What does $\log_b x = y$ mean in exponential form? | Review Logarithm Definition |
| What are the cancellation equations $\log_b(b^x) = x$ and $b^{\log_b x} = x$? | Review Logarithm Definition |
| What is the approximate numerical value of $e$? | It's approximately 2.71828... |
| What is the domain of $\log_b x$? | $(0, \infty)$: positive numbers only |
For any base $b > 0$, $b \neq 1$:
$$\log_b x = y \quad \Longleftrightarrow \quad b^y = x$$
The logarithm answers: "What power of the base gives me $x$?"
The number $e$ is defined as:
$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828...$$
It arises naturally in compound interest, population growth, and is the unique base where $\frac{d}{dx}[b^x] = b^x$.
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Course | MATH162 |
| Section | Stewart 6.3 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
The natural logarithm is the logarithm with base $e$:
$$\boxed{\ln x = \log_e x}$$
By the definition of logarithm:
$$\boxed{\ln x = y \quad \Longleftrightarrow \quad e^y = x}$$
In words: $\ln x$ is the power to which $e$ must be raised to get $x$.
Since $\ln x$ and $e^x$ are inverses:
$$\boxed{\ln(e^x) = x \quad \text{for all real } x}$$
$$\boxed{e^{\ln x} = x \quad \text{for all } x > 0}$$
These are your main tools for solving equations involving $e$ and $\ln$.
| Expression | Value | Reason |
|---|---|---|
| $\ln 1$ | $0$ | $e^0 = 1$ |
| $\ln e$ | $1$ | $e^1 = e$ |
| $\ln e^2$ | $2$ | Cancellation: $\ln(e^2) = 2$ |
| $\ln \sqrt{e}$ | $\frac{1}{2}$ | $\sqrt{e} = e^{1/2}$, so $\ln(e^{1/2}) = \frac{1}{2}$ |
| $\ln \frac{1}{e}$ | $-1$ | $\frac{1}{e} = e^{-1}$, so $\ln(e^{-1}) = -1$ |
| Property | $e^x$ | $\ln x$ |
|---|---|---|
| Domain | $\mathbb{R}$ | $(0, \infty)$ |
| Range | $(0, \infty)$ | $\mathbb{R}$ |
| $y$-intercept | $(0, 1)$ | None |
| $x$-intercept | None | $(1, 0)$ |
| Asymptote | Horizontal: $y = 0$ | Vertical: $x = 0$ |
y y
| y = e^x | y = ln(x)
| / | ___
3 + / 2 + ___/
| / | ___/
1 +----+/ 0 +---+----------→ x
| / | /|
| / | / | (1, 0)
| / -2+/ |
+/------------→ x | |
1 2 | x = 0 (asymptote)Key observations:
$$\lim_{x \to \infty} \ln x = \infty \qquad \text{(but grows very slowly)}$$
$$\lim_{x \to 0^+} \ln x = -\infty \qquad \text{(vertical asymptote)}$$
How slowly does $\ln x$ grow?
| $x$ | $\ln x$ (approx.) |
|---|---|
| $10$ | $2.3$ |
| $100$ | $4.6$ |
| $1{,}000$ | $6.9$ |
| $1{,}000{,}000$ | $13.8$ |
To get $\ln x = 20$, you need $x = e^{20} \approx 485{,}000{,}000$. That's nearly half a billion!
This identity connects powers to exponentials:
$$\boxed{x^r = e^{r \ln x} \quad \text{for } x > 0}$$
Why it works: Start with the second cancellation equation $e^{\ln x} = x$. Raise both sides to the power $r$:
$$(e^{\ln x})^r = x^r$$ $$e^{r \ln x} = x^r$$
This identity is crucial for logarithmic differentiation (Section 6.4).
Any logarithm can be expressed in terms of natural logarithms:
$$\boxed{\log_b x = \frac{\ln x}{\ln b}}$$
Why it works: Let $y = \log_b x$. Then $b^y = x$. Take natural log of both sides:
$$\ln(b^y) = \ln x$$ $$y \ln b = \ln x$$ $$y = \frac{\ln x}{\ln b}$$
Practical use: Your calculator has an $\ln$ button but probably not a $\log_7$ button. To compute $\log_7 50$:
$$\log_7 50 = \frac{\ln 50}{\ln 7} \approx \frac{3.912}{1.946} \approx 2.01$$
Strategy: Use the inverse relationship. Apply $e^{(\cdot)}$ to "undo" $\ln$, or apply $\ln$ to "undo" $e^{(\cdot)}$.
Example 1: Solve $\ln x = 5$.
Apply $e^{(\cdot)}$ to both sides: $$e^{\ln x} = e^5$$ $$x = e^5$$
Example 2: Solve $e^{5-3x} = 10$.
Take natural log of both sides: $$\ln(e^{5-3x}) = \ln 10$$ $$5 - 3x = \ln 10$$ $$x = \frac{5 - \ln 10}{3}$$
| Mistake | Why It's Wrong | Correction |
|---|---|---|
| $\ln 0 = 0$ | There is no power of $e$ that equals 0 | $\ln 0$ is undefined |
| $\ln(-5)$ exists | $e^y > 0$ always, so only positive inputs | $\ln$ of negative is undefined |
| $\ln(x + y) = \ln x + \ln y$ | Logs don't distribute over addition | This is wrong; see Laws of Logarithms |
| $e^{x+y} = e^x + e^y$ | Exponentials don't distribute | Correct: $e^{x+y} = e^x \cdot e^y$ |
| Confusing $\ln x^2$ with $(\ln x)^2$ | Different operations | $\ln x^2 = 2\ln x$ but $(\ln x)^2 = (\ln x)(\ln x)$ |
Evaluate each expression without a calculator.
(a) $\ln e^7$ (b) $\ln \frac{1}{e^3}$ (c) $e^{\ln 5}$
Find each limit.
(a) $\displaystyle\lim_{x \to 0^+} \ln(\tan^2 x)$
(b) $\displaystyle\lim_{x \to 3^+} \ln(x - 3)$
Solve for $x$.
(a) $e^{2x+1} = 15$
(b) $\ln(3x - 1) = 4$
(a) Use the change of base formula to evaluate $\log_8 5$ correct to four decimal places.
(b) Express $\log_a b \cdot \log_b a$ as a single number (no logarithms in your final answer).
Consider the function $f(x) = \ln(x - 2) - 1$.
(a) Find the domain of $f$.
(b) Find the $x$-intercept.
(c) Find the vertical asymptote.
(d) Describe how the graph of $f$ is obtained from the graph of $y = \ln x$ using transformations.
$e^x$ and $\ln x$ are Opposites:
Think of $e^x$ as a "growth machine" and $\ln x$ as an "un-growth machine."
When you compose them, they cancel:
This is the essence of inverse functions.
Looking back:
Looking ahead:
Why "natural"?
The natural logarithm appears everywhere in calculus:
No other base produces formulas this clean.
| Previous | Up | Next |
|---|---|---|
| Logarithm Definition | Skills Index | Laws of Logarithms |
Last updated: 2026-01-23