Mathematics/Lessons/Exponentials & Logarithms/Natural Logarithms
Pure · Exponentials & Logarithms · Year 1

Natural Logarithms

One base gets a name of its own: $e$. Its logarithm, $\ln$, is the tool you'll reach for constantly — it undoes $e^x$ directly, and every exam question with $e$ in it expects you to know it cold.

9 min read 4 worked examples

What ln means

The natural logarithm, written $\ln$, is simply the logarithm base $e$:

$\ln x = \log_e x \quad\Longleftrightarrow\quad \ln x = y \iff e^y = x.$

Every rule you already know for logarithms applies to $\ln$ unchanged — it's the base $10$ of the previous lesson, swapped for base $e$. What makes it worth a lesson of its own is how often $e$ turns up: exponential models, compound growth, and (later) calculus all produce $e^x$ naturally, so $\ln$ becomes the standard way of "undoing" it.

ln and eˣ cancel each other

Because $\ln$ and $e^x$ are inverse functions, composing them in either order just returns what you started with:

$\ln(e^{x}) = x \quad \text{for all } x, \qquad\qquad e^{\ln x} = x \quad \text{for } x > 0.$

These two identities are the main computational tool of this lesson. Whenever an $\ln$ sits directly outside a power of $e$ (or an $e$ sits directly outside an $\ln$), it cancels immediately — no calculator, no laws, just the definition of an inverse.

Why this cancellation is worth memorising

This is exactly the same idea as $\sqrt{x^2} = x$ or $(\sqrt{x})^2 = x$ — a function undoing its own inverse. The reason it matters so much here is practical: any equation with the unknown trapped inside a power of $e$ can be freed by taking $\ln$ of both sides, and any equation with the unknown trapped inside an $\ln$ can be freed by raising $e$ to both sides. That's the entire strategy behind solving equations involving $e$ — the next lesson builds on precisely this.

Special values and the domain of ln

Three values fall straight out of the definition, exactly as they did for a general base: $\ln e = 1$ (since $e^1 = e$), $\ln 1 = 0$ (since $e^0 = 1$), and $\ln\!\left(\dfrac1e\right) = -1$ (since $e^{-1} = \dfrac1e$).

Because $e^y$ is only ever positive whatever $y$ is, $\ln x$ is only defined for $x > 0$ — there is no real value of $\ln 0$ or $\ln(\text{negative number})$. Graphically, $y = \ln x$ passes through $(1, 0)$, climbs slowly and without bound as $x$ increases, and has a vertical asymptote at $x = 0$ — the mirror image of $y = e^x$'s horizontal asymptote at $y = 0$, reflected in the line $y = x$.

The laws of ln

Because $\ln$ is a logarithm like any other, the product, quotient and power laws carry straight across:

Product law: $\ln(xy) = \ln x + \ln y$
Quotient law: $\ln\!\left(\dfrac{x}{y}\right) = \ln x - \ln y$
Power law: $\ln\!\left(x^k\right) = k\ln x$

Nothing new to learn here beyond swapping the base — but combined with the cancellation identities above, $\ln$ becomes a genuinely powerful simplifying tool.

Worked example 1

Cancel ln and eˣ directly
Simplify (a) $\ln(e^{7})$, (b) $e^{\ln 12}$, and (c) $\ln\!\left(e^{3}\right) + e^{\ln 2}$.
MethodSpot ln directly outside e, or e directly outside ln
1
(a) $\ln(e^7)$: $\ln$ and $e$ cancel, leaving the exponent. Answer: $7$.
$\ln(e^x)=x$.
2
(b) $e^{\ln 12}$: $e$ and $\ln$ cancel, leaving what was inside. Answer: $12$.
$e^{\ln x}=x$.
3
(c) $\ln(e^3) + e^{\ln 2} = 3 + 2 = 5$.
Both cancel independently.
▪ solved
Examiner's eye

These cancellations are instant once you spot the pattern — no laws, no calculator. The only way to get them wrong is overthinking: if you see $\ln$ wrapped directly around $e^{(\text{something})}$, or $e$ wrapped directly around $\ln(\text{something})$, the answer is just that "something".

Worked example 2

Combine using the laws · single logarithm
Write $2\ln 3 + \ln 5 - \ln 15$ as a single natural logarithm, then evaluate it exactly.
MethodPower law first, then product and quotient
1
Power law: $2\ln 3 = \ln 3^2 = \ln 9$.
Coefficient becomes a power.
2
Product law: $\ln 9 + \ln 5 = \ln 45$.
Adding logs multiplies.
3
Quotient law: $\ln 45 - \ln 15 = \ln\!\left(\dfrac{45}{15}\right) = \ln 3$.
Subtracting logs divides.
4
$\ln 3$ has no simpler exact form — it's already the answer, $\approx 1.099$.
Not every ln simplifies to an integer.
▪ solved

Worked example 3

Solve a one-step equation · undo ln or e
Solve (a) $\ln x = 4$ exactly, and (b) $e^{x} = 9$, giving your answer to 3 significant figures.
MethodRaise e to both sides, or take ln of both sides
1
(a) Raise $e$ to both sides: $e^{\ln x} = e^4$, so $x = e^4$ exactly.
$e^{\ln x}=x$ frees x.
2
(b) Take $\ln$ of both sides: $\ln(e^x) = \ln 9$, so $x = \ln 9$.
$\ln(e^x)=x$ frees x.
3
Evaluate: $x = \ln 9 \approx 2.20$ (3 s.f.).
Calculator's ln button.
▪ solved
The key insight

Notice the symmetry: an equation of the form $\ln(\ldots) = \text{number}$ is undone by applying $e$ to both sides; an equation of the form $e^{(\ldots)} = \text{number}$ is undone by applying $\ln$ to both sides. You always apply the opposite operation to strip away the one that's there. This is the whole idea the next lesson extends to harder, multi-step equations.

Worked example 4

A domain check · ln of a bracket
Solve $\ln(x - 2) = 3$, and explain why the solution is valid.
MethodRaise e to both sides, then check the domain
1
Raise $e$ to both sides: $e^{\ln(x-2)} = e^{3}$, so $x - 2 = e^3$.
$e^{\ln(\ldots)}$ cancels.
2
Rearrange: $x = e^{3} + 2 \approx 22.1$ (3 s.f.).
Isolate x.
3
Check: $\ln(x-2)$ needs $x - 2 > 0$, i.e. $x > 2$. Since $22.1 > 2$, the solution is valid.
ln is only defined for a positive argument.
▪ solved
Why the check matters

Whenever the unknown sits inside the argument of a $\ln$ (rather than being the whole equation), always check that your final answer keeps that argument positive. Here it clearly does — but with more complicated equations later, a solution can come out algebraically correct and still be rejected because it makes the original $\ln$ undefined.

Common mistake

$\ln(x+y) \neq \ln x + \ln y$ — there's no law for the log of a sum, only of a product. And $e^{x+y} = e^x e^y$ (indices add), but $e^{x+y} \neq e^x + e^y$. Keep "what's inside one ln/e" separate from "two lns/e's combined", exactly as with any other base.

Check yourself

Solve $\ln(2x + 1) = 2$, giving your answer to 3 significant figures, and confirm the solution is valid.
(Hint: raise e to both sides first, then check $2x+1 > 0$.)

Method
1
Raise $e$ to both sides: $2x + 1 = e^{2}$.
Cancels the ln.
2
$2x = e^2 - 1$, so $x = \dfrac{e^2 - 1}{2} \approx 3.19$ (3 s.f.).
Isolate x.
3
Check: $2x+1 = e^2 \approx 7.39 > 0$, so the solution is valid.
The ln's argument stays positive.
▪ solved

Raise $e$ to both sides to strip the $\ln$ immediately, solve the resulting linear equation as normal, then always confirm the argument of the original $\ln$ is positive.

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