Why does e^(ln x) = x need x > 0 but ln(eˣ) = x works for all x?

The round trip x → ln x → eˣ starts in ln's domain, which is only positive numbers. The trip x → eˣ → ln x starts in eˣ's domain (all reals), and eˣ is always positive, so ln accepts it. ln is the picky one.

Is ln(e^(−2)) really −2?

Yes. ln(eˣ) = x for every real x, including negatives: e^(−2) ≈ 0.135 is positive, so ln accepts it, and ln undoes the e^ to return −2.

Calculus I › Exponential & log laws › full formula sheet

e^{ln x} = x · ln(e^x) = x

Say it: “e to the x and the natural log of x undo each other”

Inverse pair

Q: What is the inverse-pair law?

eˣ and ln x undo each other: e^(ln x) = x and ln(eˣ) = x. They are inverse functions — each one reverses what the other does.

Q: How do you solve e^(2x) = 7?

Take ln of both sides: ln(e^(2x)) = ln 7 collapses to 2x = ln 7, so x = (ln 7)/2. The inverse pair does the undoing.

Q: How do you solve ln x = 4?

Raise e to both sides: e^(ln x) = e⁴ collapses to x = e⁴. Same pair, opposite direction.

e^x and ln x undo each other — the key that unlocks every equation in this chapter.

Notation on this page: ln is the natural log (base e). Watch the domains: the two halves of this pair are not symmetric.

Before this lesson: Definition of e

Where it comes from

Equations like e^3x = 10 or ln x = 2 trap x where algebra can't reach it — inside an exponent, inside a log. You need a function that undoes e^x and ln. The lazy move is to pile on more of the same:

e 3x = 10 \Rightarrow e (e 3x) = e 10 ?? the tempting — and wrong — guess: "take e^ of both sides"

That compounds the problem instead of undoing it. The right undo is the inverse function: ln undoes e^x, e^x undoes ln. Here is why, from the definition. ln x is defined as "the power you raise e to in order to get x":

ln 7 ≈ 1.946

means

e^1.946 ≈ 7

ln 7 is the exponent that turns e into 7 — that is the definition, not a theorem.

e^{ln 7}

e^1.946 ≈ 7

So e^{(the exponent that gives 7)} = 7. They eat each other: e^{ln x} = x.

ln(e²)

Reverse direction: ln asks "what power of e gives e²?" — the power 2. So ln(e^x) = x.

This pair is the hidden engine of the whole topic: every log-law derivation on the previous pages used it (Step 1 and Step 3 of each proof). Learn it cold and the other five laws become corollaries.

Before reading on: ln 7 is defined as “the exponent that turns e into 7.” So what must e^{ln 7} equal — and why is there nothing to compute?

Derivation

There is almost nothing to prove — the pair is the definition of ln as the inverse of exp. The real content is the domain asymmetry: the two round trips are not identical.

ln(e^x)

x (all real x)

Step 1 — by definition. ln is defined as the inverse of e^x: it answers "which exponent gives this value?" For e^x the answer is x itself — no work needed.

e^{ln x}

x (x > 0 only)

Step 2 — the onto argument. e^x maps ℝ onto (0,∞): every positive x equals e^y for some y, and that y is ln x. So e^{ln x} = e^y = x — but only for x > 0, since ln x doesn't exist otherwise.

x → ln x → e^{( )}

x → e^x → ln( )

Step 3 — key step: why they differ. The first trip starts in ln's domain — positives only. The second trip starts in e^x's domain (all reals), and e^x always lands positive, so ln happily accepts it. ln is the picky one. ∎

The negative test. ln(e⁻²) = −2 works fine — e⁻² ≈ 0.135 is positive, so ln accepts it. But e^ln(−2) is undefined — the trip dies at the first stop. When in doubt, test with −2: it separates the two halves instantly.

Before reading on: e^ln(−2): does the pair collapse here? What breaks first?

How to use it

The procedure, every time:

Spot the wrapping: e^{ln(something)} or ln(e^something). The functions are nested — that is the signal.
Check the domain. e^{ln x} = x needs x > 0. ln(e^x) = x needs nothing — all real x.
Collapse: e^ln(3x) = 3x (x > 0); ln(e^x²) = x² (all x).
Solve equations by undoing. See e^(…) = number → take ln of both sides. See ln(…) = number → raise e to both sides.

The equation-solving reflex

e^2x = 5

⇒

ln(e^2x) = ln 5 ⇒ 2x = ln 5 ⇒ x = (ln 5)/2

ln undoes e^x: the left side collapses to 2x.

ln x = 4

⇒

e^{ln x} = e⁴ ⇒ x = e⁴

e^x undoes ln: the left side collapses to x.

Tricky cases

e^ln(x²) = x² — fine for all x ≠ 0 (x² > 0, so ln accepts it). e^{2·ln x} = e^ln(x²) = x² (x > 0) — the pair teams up with the power law. ln(e^x + 1) does not collapse — the +1 means it is not a pure ln(e^something). The wrapping must be exact.

Common mistake: writing e^ln(x−1) = x−1 with no domain note, then plugging x = 0 to get e^ln(−1). The collapse is valid, but only where ln(x−1) exists: x > 1. State the domain every time.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic collapse: e^{ln 7}

Spot the wrapping. e^{( )} around ln 7 — the pair, e-outside. ✓
Check the domain. ln 7 needs 7 > 0. ✓
Collapse: = 7. (Why? ln 7 is the exponent that gives 7; raising e to it returns 7.)
Check numerically. ln 7 ≈ 1.9459; e^1.9459 ≈ 7. ✓

Common mistake: e^{ln 7} = e · ln 7 ≈ 5.29 — reading the exponent as multiplication. e^{ln 7} is e raised to ln 7, and the pair collapses it to 7 exactly.

Your turn: Simplify e^{ln 5}.

Answer: 5.

The pair, e-outside: ln 5 is the exponent that gives 5, so raising e to it returns 5.

Example 2 — the reverse: ln(e⁻⁴)

Spot the wrapping. ln around e⁻⁴ — the pair, ln-outside. ✓
Check the domain. ln(e^x) = x works for all real x — including −4. (e⁻⁴ ≈ 0.018 > 0, so ln accepts it.)
Collapse: = −4.

Common mistake: declaring ln(e⁻⁴) “undefined because −4 is negative.” The −4 is the exponent, not the log's argument — the argument is e⁻⁴ ≈ 0.018, safely positive. This half of the pair never fears negatives.

Your turn: Simplify ln(e⁻¹).

Answer: −1.

The pair, ln-outside: ln(e^x) = x for every real x, including −1.

Example 3 — solving: e^3x = 10

Undo with ln. Take ln of both sides: ln(e^3x) = ln 10. (Why ln? It is the inverse of e^x — it collapses the left side.)
Collapse. 3x = ln 10.
Solve. x = (ln 10)/3 ≈ 0.768.
Check. e^3·0.768 = e^2.303 ≈ 10. ✓

Common mistake: “taking e of both sides”: e^{(e^3x)} = e¹⁰ — a tower, vastly worse. Undo with the inverse function, not the same one.

Your turn: Solve e^2x = 7.

Answer: x = (ln 7)/2 ≈ 0.973.

Take ln of both sides: 2x = ln 7, so x = (ln 7)/2. Check: e^1.9459 ≈ 7 ✓.

Example 4 — teaming up: simplify e^{3·ln x}, x > 0

Bring the multiplier inside (power law, backwards): 3·ln x = ln(x³). So e^{3·ln x} = e^ln(x³).
Collapse with the pair: e^ln(x³) = x³. (Domain: x > 0 given, so ln(x³) exists.)
Check at x = 2. e^{3·ln 2} = e^{ln 8} = 8 = 2³. ✓

The lesson: the pair rarely works alone — here it teamed with the power law. e^{(log expression)} almost always simplifies: move multipliers inside as exponents, then collapse.

Common mistake: e^{3·ln x} = 3x — “cancelling” the e and ln while ignoring the 3. The 3 must move inside the log first (as an exponent); only then does the pair collapse.

Your turn: Simplify e^{2·ln x}, x > 0.

Answer: x².

Bring the multiplier inside (power law backwards): 2 ln x = ln(x²); then e^ln(x²) = x² ✓.

Memorization tips

Say the pair: “e to the ln of x is x; ln of e to the x is x.” The rhythm matches the symmetry.
They eat each other. See e^ln(…) or ln(e^(…)) and collapse on sight — no algebra needed.
ln is the picky one: ln(e^x) = x for ALL x, but e^{ln x} = x only for x > 0. The trip that starts at ln is the restricted one.
The −2 test: ln(e⁻²) = −2 ✓ works; e^ln(−2) ✗ undefined. One number separates the halves forever.
Equation reflex: see e^(…) = number → take ln. See ln(…) = number → raise e. Undo with the inverse, never the same function.
The engine of the chapter: every log law's proof used this pair twice. If the pair is reflex, the other five laws are one step each.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the inverse pair is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the inverse-pair law?

e^x and ln x undo each other: e^{ln x} = x and ln(e^x) = x. They are inverse functions — each one reverses what the other does.

Why does e^{ln x} = x need x > 0 but ln(e^x) = x works for all x?

The round trip x → ln x → e^x starts in ln’s domain, which is only positive numbers. The trip x → e^x → ln x starts in e^x’s domain (all reals), and e^x is always positive, so ln accepts it. ln is the picky one.

How do you solve e^2x = 7?

Take ln of both sides: ln(e^2x) = ln 7 collapses to 2x = ln 7, so x = (ln 7)/2. The inverse pair does the undoing.

How do you solve ln x = 4?

Raise e to both sides: e^{ln x} = e⁴ collapses to x = e⁴. Same pair, opposite direction.

Is ln(e⁻²) really −2?

Yes. ln(e^x) = x for every real x, including negatives: e⁻² ≈ 0.135 is positive, so ln accepts it, and ln undoes the e^ to return −2.

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Inverse pair

Where it comes from

Derivation

How to use it

The equation-solving reflex

Tricky cases

Worked examples

Example 1 — the basic collapse: eln 7

Example 2 — the reverse: ln(e−4)

Example 3 — solving: e3x = 10

Example 4 — teaming up: simplify e3·ln x, x > 0

Memorization tips

Final challenge

How to learn a formula here

Frequently asked questions

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Calculus I formula sheet

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Example 1 — the basic collapse: e^{ln 7}

Example 2 — the reverse: ln(e⁻⁴)

Example 3 — solving: e^3x = 10

Example 4 — teaming up: simplify e^{3·ln x}, x > 0