Calculus I › Exponential & log laws › full formula sheet

e^x · e^y = e^x+y

Say it: “e to the x times e to the y equals e to the x plus y”

Product of powers

Q: Why isn't eˣ·eʸ equal to eˣʸ?

Test x = 2, y = 3: e²·e³ = e⁵ ≈ 148.4, but eˣʸ = e⁶ ≈ 403.4. Multiplying the exponents is the power-of-a-power law's job — it needs parentheses: (eˣ)ʸ = eˣʸ.

Q: Should I combine exponentials before differentiating?

Yes — always. d/dx[eˣ·e²ˣ] becomes d/dx[e³ˣ] = 3e³ˣ, one line instead of a full product rule. Simplifying first is the single highest-value habit in this chapter.

Q: Does the law work for bases other than e?

Yes, for any positive base: aᵐ·aⁿ = aᵐ⁺ⁿ, by the same counting argument. The bases must match — 2³·3² cannot combine.

How to multiply two exponentials — and why you add the exponents instead of multiplying them.

Notation on this page: e ≈ 2.718 is Euler's number; x and y are real numbers.

Before this lesson: Definition of e

Where it comes from

Products of exponentials are everywhere in calculus: e^2x·e^3x, e^x·e^−x, growth models multiplied together. Staring at the symbols, the tempting move is to multiply the exponents:

e x \cdot e y = e xy ?? the tempting — and wrong — guess

Kill it with numbers. Take x = 2, y = 3:

e² · e³

≈

7.389 · 20.086 ≈ 148.4

Multiplying the actual values gives about 148.4.

e²⁺³ = e⁵

≈

148.4

Adding the exponents matches the real product exactly.

e^2·3 = e⁶

≈

403.4

Multiplying the exponents gives 403.4 — not even close. Dead on arrival.

Here is the intuition that makes the real law obvious. An exponent counts repeated factors: e² means e·e (two e's), e³ means e·e·e (three e's). Multiplying concatenates the two lists:

e² · e³

(e·e) · (e·e·e)

Two e's, then three e's.

e·e·e·e·e = e⁵

Five e's total: 2 + 3 = 5. Concatenating lists adds the counts — that is the whole law.

Exponent notation is Descartes' (1637); the extension of these counting laws to real exponents — which calculus needs, since e^π is a perfectly good number — came with Euler's treatment of e^x as a function. The counting argument is the seed; the derivation below grows it to all real exponents.

Before reading on: e² · e³ means (e·e)·(e·e·e). Count the e’s — what should the exponent rule be for e^m · eⁿ?

Derivation

Start with positive integers, where "eⁿ" literally means n copies of e multiplied. Write E(x) = e^x for the real-exponent function.

e^m · eⁿ

(e·…·e) · (e·…·e)

Step 1 — what eⁿ means. For positive integers m, n: m factors, then n factors. Regrouping is just associativity and commutativity of multiplication.

e·…·e = e^m+n

Step 2 — count. There are m + n factors in the combined list. The integer case is proved — but calculus needs irrational exponents too, where you cannot count "√2 factors".

F(x) = E(x+y)/E(y)

and

F′(x) = F(x), F(0) = 1

Step 3 — key step: the extension to all reals. Fix y. Since E′ = E (the defining property of e^x), differentiating gives F′(x) = E′(x+y)/E(y) = E(x+y)/E(y) = F(x); and F(0) = E(y)/E(y) = 1.

⇒

E(x+y) = E(x)·E(y)

Step 4 — uniqueness. E itself satisfies f′ = f, f(0) = 1, and it is the only function that does. So F(x) = E(x), i.e. E(x+y)/E(y) = E(x). Multiply through: e^x+y = e^x·e^y. ∎

Any positive base works. The counting argument never used anything special about e, so a^m·aⁿ = a^m+n for any a > 0. What is special about e is the calculus: d/dx [e^x] = e^x, which is exactly the property the derivation above leaned on. The one hard requirement: the bases must match — 2³·3² = 8·9 = 72, while 6⁵ = 7776. Different objects being counted, no combining.

Before reading on: is e^2x · e^3x equal to e^6x or e^5x? Commit before you read on.

How to use it

The procedure, every time:

Confirm the same base. e^2x·e^3x ✓. e^x·2 ✗ (2 is not a power of e — nothing to combine). 2^x·3^y ✗ (different bases).
Keep the base, add the exponents. e^2x·e^3x = e^2x+3x. Add them as they are — do not multiply, do not touch the base.
Simplify the exponent. = e^5x. Combine like terms up there.
Use it before any calculus. d/dx [e^x·e^2x] becomes d/dx [e^3x] = 3e^3x — one line instead of a full product rule.

Combine first, always

This is the highest-value habit in the chapter: never differentiate or integrate a product of exponentials before combining it.

∫ e^x·e^4x dx

∫ e^5x dx

Combine first: e^x·e^4x = e^5x.

e^5x/5 + C

Done in one line.

Read it backwards: splitting

e x+y = e x \cdot e y splitting — half the uses of this law run in reverse Say it: e to the x plus y splits into e to the x times e to the y

Splitting shines in limits: e^x+1/e^x = e^x·e¹/e^x = e. And in factoring: e^2x + e^x = e^x(e^x + 1).

Tricky cases

Constants in the exponent combine too: e²·e^x = e^x+2. Cancellation: e^x·e^−x = e⁰ = 1 — if your simplification does not collapse to 1 here, something is wrong. Coefficients multiply separately from exponents: 5e^2x·2e^3x = 10e^5x (coefficients multiply, exponents add — never mix the two operations).

Common mistake: writing e^2x·e^3x = e^6x (multiplying exponents) or e^6x² (multiplying everything). The paren test: no parens around a power being raised → add. Parens like (e^x)^y → that is the next law (power of a power), and there you multiply.

Worked examples

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

Example 1 — the basic move: e³ · e⁴

Confirm the same base. Both are powers of e. ✓
Add the exponents. 3 + 4 = 7, so = e⁷.
Check numerically. e³ ≈ 20.086, e⁴ ≈ 54.598; their product ≈ 1096.6, and e⁷ ≈ 1096.6. Matches ✓

Common mistake: e^3·4 = e¹² ≈ 162755 — off by a factor of 148. Multiplying exponents is never the move without parens.

Your turn: e⁵ · e².

Answer: e⁷.

Same base: add exponents, 5 + 2 = 7. Check: e⁵ ≈ 148.4, e² ≈ 7.389, product ≈ 1096.6 = e⁷ ✓.

Example 2 — negative exponents: e^2x · e^−5x

Same base. ✓ — negative exponents follow the same law.
Add: 2x + (−5x) = −3x. So = e^−3x = 1/e^3x. (Why the reciprocal? e^−3x = 1/e^3x by the negative-exponent rule.)
Check at x = 1. e²·e⁻⁵ ≈ 7.389 · 0.006738 ≈ 0.0498, and e⁻³ ≈ 0.0498. ✓

Common mistake: writing e^2x−5x = e^−3x and then “simplifying” to −e^3x. A negative exponent means reciprocal, not a negative value — e^−3x is always positive.

Your turn: e^3x · e^−7x.

Answer: e^−4x = 1/e^4x.

Negative exponents follow the same law: 3x + (−7x) = −4x ✓.

Example 3 — combine before differentiating: f(x) = e^2x · e^3x

Path A — combine first (one line).

e^2x·e^3x = e^5x.
f′(x) = 5e^5x.

Path B — product rule (no simplifying).

f′ = 2e^2x·e^3x + e^2x·3e^3x = 2e^5x + 3e^5x = 5e^5x. Same answer ✓ — five times the writing.

The lesson: combining first is not a trick, it is the workflow. On an exam, Path A saves a minute and dodges every product-rule slip.

Common mistake: differentiating each factor and multiplying the derivatives: 2e^2x·3e^3x = 6e^5x. That is the f′g′ error wearing an exponential disguise.

Your turn: Differentiate f(x) = e^x · e^4x by combining first.

Answer: f′(x) = 5e^5x.

e^x·e^4x = e^5x, so f′(x) = 5e^5x — one line, no product rule.

Example 4 — solving: e^2x · e^x−1 = e⁵

Combine the left side. e^2x·e^x−1 = e^2x+x−1 = e^3x−1.
Equate exponents. e^3x−1 = e⁵. (Why allowed? e^x is strictly increasing, hence one-to-one: equal outputs force equal inputs.) So 3x − 1 = 5.
Solve. 3x = 6, so x = 2.
Check. e⁴·e¹ = e⁵. ✓

Common mistake: taking logs too early and writing 2x·ln e + (x−1)·ln e = 5·ln e. That works but skips the combine step — and on messier equations the uncombined form is where sign errors breed.

Your turn: Solve e^x · e^2x−3 = e⁷.

Answer: x = 10/3.

Left side: e^3x−3 = e⁷; e^x is one-to-one, so 3x − 3 = 7 and x = 10/3 ✓.

Memorization tips

Say it aloud: “same base — add the exponents.” One sentence, the whole law.
Count, don't compute: e²·e³ is two e's then three e's — five e's. The picture is the proof for integers.
The cancellation test (your 5-second self-check): e^x·e^−x must collapse to e⁰ = 1. If your rule doesn't do that, it's the wrong rule.
Don't cross with the next law: product of powers ADDS (no parens); power of a power MULTIPLIES (parens). The parens are the entire difference.
Read it backwards: e^x+y → e^x·e^y. Splitting is half of this law's uses — limits, factoring, separating constants.
Numeric anchor: e²·e³ ≈ 148.4 = e⁵. If you ever blank on add-vs-multiply, e⁶ ≈ 403.4 is obviously too big.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the product of powers 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 product-of-powers law?

When you multiply two powers of the same base, keep the base and add the exponents: e^x·e^y = e^x+y. For example, e²·e³ = e⁵.

Why isn’t e^x·e^y equal to e^xy?

Test x = 2, y = 3: e²·e³ = e⁵ ≈ 148.4, but e^xy = e⁶ ≈ 403.4. Multiplying the exponents is the power-of-a-power law’s job — it needs parentheses: (e^x)^y = e^xy.

Why do the exponents add?

An exponent counts repeated factors: e² = e·e, e³ = e·e·e. Multiplying concatenates the two factor lists, so the counts add: 2 + 3 = 5 factors, e⁵.

Should I combine exponentials before differentiating?

Yes — always. d/dx [e^x·e^2x] becomes d/dx [e^3x] = 3e^3x, one line instead of a full product rule. Simplifying first is the single highest-value habit in this chapter.

Does the law work for bases other than e?

Yes, for any positive base: a^m·aⁿ = a^m+n, by the same counting argument. The bases must match — 2³·3² cannot combine.

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Product of powers

Where it comes from

Derivation

How to use it

Combine first, always

Read it backwards: splitting

Tricky cases

Worked examples

Example 1 — the basic move: e³ · e⁴

Example 2 — negative exponents: e2x · e−5x

Example 3 — combine before differentiating: f(x) = e2x · e3x

Example 4 — solving: e2x · ex−1 = e⁵

Memorization tips

Final challenge

How to learn a formula here

Frequently asked questions

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Example 2 — negative exponents: e^2x · e^−5x

Example 3 — combine before differentiating: f(x) = e^2x · e^3x

Example 4 — solving: e^2x · e^x−1 = e⁵