How is (eˣ)ʸ different from eˣ·eʸ?

The parentheses are the entire difference. eˣ·eʸ has no outer exponent, so the exponents add (eˣ⁺ʸ). (eˣ)ʸ raises the whole power again, so the exponents multiply (eˣʸ).

Calculus I › Exponential & log laws › full formula sheet

(e^x)^y = e^xy

Say it: “e to the x, all raised to the y, equals e to the x times y”

Power of a power

Q: Why do the exponents multiply?

(eˣ)ʸ is y copies of eˣ multiplied together. The product-of-powers law adds the exponent y times: x + x + … + x = xy.

Q: What is (e^(2x))^(1/2)?

eˣ. Multiply: 2x·(1/2) = x. There is no ± ambiguity because e^(2x) is always positive.

Raising a power to a power: multiply the exponents — and how to stop mixing it up with the product-of-powers law.

Notation on this page: e ≈ 2.718 is Euler's number; x and y are real numbers. Compare with the previous law: e^x·e^y = e^x+y (no parens, add).

Before this lesson: Product of powers

Where it comes from

Nested powers like (e^2x)³ show up constantly — and students fresh off the product-of-powers law do the natural thing and add:

(e x) y = e x+y ?? the tempting — and wrong — guess

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

(e²)³

e² · e² · e² = e²⁺²⁺² = e⁶

Cubing means three copies multiplied: e²·e²·e². The product-of-powers law adds: 2+2+2 = 6. Value ≈ 403.4.

e²⁺³ = e⁵

≈

148.4

The naive add-guess gives 148.4 — less than half the true value. Dead on arrival.

Here is the intuition that makes the real law obvious. (e^x)^y is y copies of e^x multiplied together. We already know what to do with a product of same-base powers — the previous law adds the exponents. Adding x to itself y times is multiplication:

(e^x)^y

e^x · e^x ·…· e^x (y copies)

That is what raising to the y-th power means.

e^x+x+…+x = e^xy

Product-of-powers adds the exponent y times: x + … + x = xy. The new law is the old law, applied repeatedly.

So the two laws are not rivals — this one contains the previous one. The parentheses are the entire difference: no outer parens, add; outer parens, multiply.

Before reading on: (e²)³ means e² · e² · e². What does the product-of-powers law turn that into — and what’s the general pattern?

Derivation

For a positive integer n, (e^x)ⁿ is n copies multiplied — then the product-of-powers law does the rest. Real exponents need one definition first: for a > 0, a^y := e^y·ln a.

(e^x)ⁿ

e^x · e^x ·…· e^x (n copies)

Step 1 — what ( )ⁿ means. An integer exponent is repeated multiplication. This step uses the definition of ^n, not any law.

e^x+x+…+x = e^nx

Step 2 — the product-of-powers law. n copies of e^x multiplied: add the exponent n times. Integer case proved.

(e^x)^y

e^{y·ln (e^x)} = e^y·x = e^xy

Step 3 — key step: real y. Apply the definition a^y := e^y·ln a with a = e^x, then collapse ln (e^x) = x — the inverse pair. That one cancellation is the whole extension. ∎

The tower trap. (e^x)^y is not e^{(x^y)}. Compare: (e²)³ = e⁶ ≈ 403.4, but e^(2³) = e⁸ ≈ 2981. Exponent towers evaluate top-down; parentheses change the meaning entirely. If you see a stacked exponent with no parens, it is a tower, not this law.

Before reading on: which is bigger: (e³)² or e³ · e²? Compute it both ways.

How to use it

The procedure, every time:

Spot the shape: a power raised to a power — parens with an exponent outside: (e^2x)³. No outer parens? Then it is the product-of-powers law (add), not this one.
Multiply the exponents, keep the base. (e^2x)³ = e^6x. Multiply everything upstairs: 2x · 3 = 6x.
Simplify before calculus. d/dx [(e^3x)²] becomes d/dx [e^6x] = 6e^6x — one line instead of a nested chain rule.

The paren test

e x \cdot e y = e x+y vs (e x) y = e xy no parens \to add \cdot parens \to multiply Say it: no parens means add the exponents; parens around the power means multiply them

Tricky cases

Fractional exponents: (e^2x)^1/2 = e^x. No ± ambiguity — e^2x is always positive, so the principal root is the only root. Negative exponents: (e^x)⁻¹ = e^−x = 1/e^x. Nested: ((e^x)²)³ = e^6x — work inside-out, or just multiply all three: x·2·3. Half-powers in reverse: e^x/2 = (e^x)^1/2 = √(e^x).

Common mistake: writing (e^2x)³ = e^2x³ — stacking the exponents into a tower instead of multiplying. e^2x³ means e raised to 2x³, a completely different (and much bigger) function. Parens mean multiply: 2x·3 = 6x.

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)²

Spot the shape. Parens with an exponent outside — power of a power. ✓
Multiply: 3 · 2 = 6, so = e⁶.
Check numerically. e³ ≈ 20.086; squared ≈ 403.43; e⁶ ≈ 403.43. ✓

Common mistake: e³⁺² = e⁵ ≈ 148.4 — adding, the product-of-powers reflex. The parens are the signal: multiply.

Your turn: (e⁴)³.

Answer: e¹².

Multiply the exponents: 4 · 3 = 12. Check: e⁴ ≈ 54.6, cubed ≈ 162755 = e¹² ✓.

Example 2 — with variables: (e^2x)³

Shape: power raised to a power. ✓
Multiply everything upstairs: (2x)·3 = 6x. So = e^6x.
Check at x = 1. (e²)³ = e⁶ from Example 1; formula gives e⁶. ✓

Common mistake: e^2x³ — multiplying only the 3 into the exponent's coefficient position wrong, i.e. stacking. Multiply the whole exponent: (2x)·3, not 2·(x³) and not 2x³.

Your turn: (e^3x)².

Answer: e^6x.

Parens with an exponent outside: multiply everything upstairs, (3x)·2 = 6x ✓.

Example 3 — simplify before differentiating: d/dx [(e^x)²]

Path A — simplify first (one line).

(e^x)² = e^2x.
d/dx [e^2x] = 2e^2x.

Path B — chain rule directly (no simplifying).

d/dx [(e^x)²] = 2(e^x)·e^x = 2e^2x. Same answer ✓ — but the unsimplified form is where students drop the inner e^x.

The lesson: simplifying first turns a two-layer chain rule into a one-layer one. Fewer layers, fewer dropped factors.

Common mistake: writing d/dx [(e^x)²] = 2e^x — differentiating the outside and forgetting the inside's derivative (another e^x). Path A makes this error impossible.

Your turn: Differentiate d/dx [(e^x)³] by simplifying first.

Answer: 3e^3x.

(e^x)³ = e^3x, so the derivative is 3e^3x — no chain-rule nesting to fumble.

Example 4 — solving: (e^2x)³ = e¹²

Collapse the left side. (e^2x)³ = e^6x.
Equate exponents. e^6x = e¹² ⇒ 6x = 12 (e^x is one-to-one).
Solve. x = 2.
Check. (e⁴)³ = e¹². ✓

Common mistake: “distributing” the cube as (e^2x)³ = e^2x·e^2x·e^2x and then stopping — that is correct but unfinished; the product-of-powers law finishes it to e^6x. Don't stop halfway.

Your turn: Solve (e^3x)² = e¹⁸.

Answer: x = 3.

Left side collapses to e^6x = e¹⁸; e^x is one-to-one, so 6x = 18 and x = 3 ✓.

Memorization tips

Say the pair aloud: “product of powers adds; power of a power multiplies.” The parens tell you which.
Three copies: (e²)³ is three copies of e², each bringing its 2 along: 2+2+2 = 6 = 2·3. Repeated addition is multiplication.
The tower guard: (e²)³ = e⁶ but e^(2³) = e⁸. Stacked exponents without parens evaluate top-down — a different beast.
Self-check: (e¹)² = e² is obvious and anchors multiply. If your rule gives e³ here, it's the wrong rule.
Read it backwards: e^xy → (e^x)^y. Splitting a product-exponent is how half-powers appear: e^x/2 = √(e^x).
Numeric anchor: (e²)³ ≈ 403.4. Adding gives 148.4 (too small), tower gives 2981 (too big) — 403.4 sits between, exactly where e⁶ lives.

Final challenge

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

When a power is raised to another power, keep the base and multiply the exponents: (e^x)^y = e^xy. For example, (e²)³ = e⁶.

How is (e^x)^y different from e^x·e^y?

The parentheses are the entire difference. e^x·e^y has no outer exponent, so the exponents add (e^x+y). (e^x)^y raises the whole power again, so the exponents multiply (e^xy).

Why do the exponents multiply?

(e^x)^y is y copies of e^x multiplied together. The product-of-powers law adds the exponent y times: x + x + … + x = xy.

Is (e^x)^y the same as e^{(x^y)}?

No — this is the tower trap. (e²)³ = e⁶, but e^(2³) = e⁸. Exponent towers evaluate top-down, and the parentheses change the meaning entirely.

What is (e^2x)^1/2?

e^x. Multiply: 2x·(1/2) = x. There is no ± ambiguity because e^2x is always positive.

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Power of a power

Where it comes from

Derivation

How to use it

The paren test

Tricky cases

Worked examples

Example 1 — the basic move: (e³)²

Example 2 — with variables: (e2x)³

Example 3 — simplify before differentiating: d/dx [(ex)²]

Example 4 — solving: (e2x)³ = e12

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 2 — with variables: (e^2x)³

Example 3 — simplify before differentiating: d/dx [(e^x)²]

Example 4 — solving: (e^2x)³ = e¹²