Calculus I › Exponential & log laws › full formula sheet
Say it: “the natural log of a times b equals the natural log of a plus the natural log of b”
Log of a product
The log turns multiplication into addition — the mirror image of ex turning addition into multiplication.
Notation on this page: ln is the natural log (base e); a and b are positive real numbers.
Before this lesson: Product of powers · Inverse pair
Where it comes from
Logs of messy products — ln(2x), ln(x(x+1)) — are unpleasant to differentiate as-is. The instinct is to distribute the log the way you would distribute multiplication:
Kill it with the simplest possible numbers: a = b = e.
And while we are killing guesses, kill the other classic too — the one that costs more exam marks than any other log error:
Here is the intuition that makes the real law obvious. You already know ex turns addition into multiplication: ex+y = ex·ey. Now ln is the inverse of ex — it runs the same road backwards. So of course it must turn multiplication back into addition:
This mirror is not a metaphor — it is literally how slide rules worked for 300 years: adding lengths on the rule multiplied the numbers. The derivation below makes the mirror rigorous.
Before reading on: ex+y = ex · ey turns addition into multiplication. If ln runs that same road backwards, what must ln(a·b) become?
Derivation
Let a, b > 0. Write each as a power of e — which we can do because ln is the inverse of ex — then let the product-of-powers law do the work.
Domain note. The law as stated needs a, b > 0, because ln of a nonpositive number is undefined (in the reals). What about ln((−2)(−3)) = ln 6? That works — but you cannot split it as ln(−2) + ln(−3). The safe general form uses absolute values: ln|ab| = ln|a| + ln|b|. In calculus problems the factors are usually positive anyway; when they aren't, reach for the absolute values.
Before reading on: ln(2+3): can you split it into ln 2 + ln 3? Run the numbers before you read the verdict.
How to use it
The procedure, every time:
- Confirm it is a log of a product. ln(2x) ✓. ln(x+2) ✗ — a sum, never splits. ln x + 2 ✗ — nothing inside to split.
- Check positivity. Every factor must be positive (or use |·|). ln(2x) needs x > 0.
- Split: ln(2x) = ln 2 + ln x. One log becomes a sum of simpler logs.
- Use it before calculus. d/dx [ln(2x)] becomes d/dx [ln 2 + ln x] = 0 + 1/x = 1/x — the constant vanishes and the chain rule never appears.
Read it backwards: combining
Combining shines when a sum of logs is ugly: ln(x²) + ln(x³) collapses to ln(x⁵) before you differentiate, and in series where ln 2 + ln 3 + … telescopes into a single log. Also: ln(x(x−2)) expands to ln x + ln(x−2) for x > 2.
Split vs. chain rule
d/dx [ln(4x)]: split first → d/dx [ln 4 + ln x] = 1/x. Chain rule directly: (1/(4x))·4 = 1/x — same answer, more moving parts. Splitting wins whenever a factor is constant.
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: expand ln(10x), x > 0
- Confirm: log of a product, both factors positive. ✓
- Split: ln(10x) = ln 10 + ln x.
- Check at x = 5. ln 50 ≈ 3.912; ln 10 + ln 5 ≈ 2.303 + 1.609 = 3.912. ✓
Your turn: Expand ln(7x), x > 0.
Answer: ln 7 + ln x.
Log of a product with both factors positive — split it ✓.
Example 2 — combining: ln 4 + ln 5
- Spot the shape: a sum of two logs — the law read backwards.
- Combine: ln 4 + ln 5 = ln(4·5) = ln 20. (Why? The law says ln(ab) = ln a + ln b; running it in reverse is always legal.)
- Check. ln 4 + ln 5 ≈ 1.386 + 1.609 = 2.996; ln 20 ≈ 2.996. ✓
Your turn: Combine ln 2 + ln 8.
Answer: ln 16.
The law read backwards: ln(2·8) = ln 16. Check: 0.693 + 2.079 = 2.773 = ln 16 ✓.
Example 3 — differentiate smart: d/dx [ln(x·ex)]
- Split first: ln(x·ex) = ln x + ln(ex) = ln x + x. (ln(ex) = x — the inverse pair.)
- Differentiate the sum: d/dx [ln x + x] = 1/x + 1.
- Check via chain rule. d/dx [ln(x·ex)] = (1/(x·ex))·(ex + x·ex) = (1+x)/x = 1/x + 1. ✓ Same — but the chain-rule path is where the algebra slips happen.
Your turn: Differentiate d/dx [ln(3x²)], x > 0.
Answer: 2/x.
Split first: ln 3 + 2 ln x — the constant dies, giving 0 + 2/x. (Chain-rule check: (1/(3x²))(6x) = 2/x ✓.)
Example 4 — solving: ln x + ln(x−1) = ln 6
- Combine the left side. ln x + ln(x−1) = ln(x(x−1)) = ln(x²−x).
- Drop the logs. ln A = ln B ⇒ A = B (ln is one-to-one): x² − x = 6.
- Solve. x² − x − 6 = 0 ⇒ (x−3)(x+2) = 0, so x = 3 or x = −2.
- Check the domain — reject x = −2. ln(−2) is undefined, and x−1 = −3 is also negative. Only x = 3 survives: ln 3 + ln 2 = ln 6. ✓
Your turn: Solve ln x + ln(x−2) = ln 8.
Answer: x = 4.
Combine: ln(x(x−2)) = ln 8, so x² − 2x = 8, i.e. (x−4)(x+2) = 0. Reject x = −2 (domain); x = 4 works: ln 4 + ln 2 = ln 8 ✓.
Memorization tips
- Say it aloud: “log turns times into plus.” Four words, the whole law.
- The mirror: ex turns + into ×; ln turns × into +. They are inverses, so of course they mirror — learn them as one fact, not two.
- The slide-rule image: for 300 years, people multiplied by adding lengths on a stick. Adding lengths = adding logs = multiplying numbers. That stick was this law.
- Self-check: ln(e·e) = ln(e²) = 2 = 1 + 1. If your rule doesn't give 2 here, it's wrong.
- The sum guard: ln(a+b) NEVER splits. Test with 2+3: ln 5 ≈ 1.61 vs ln 2 + ln 3 ≈ 1.79. Tattoo this one.
- Split before calculus: constants vanish after splitting (d/dx [ln 2] = 0). Every constant factor you split off is one fewer chain-rule term.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the log of a product 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 log-of-a-product law?
The log of a product is the sum of the logs: ln(ab) = ln a + ln b (for a, b > 0). For example, ln 6 = ln 2 + ln 3.
Why isn’t ln(ab) equal to (ln a)(ln b)?
Test a = b = e: ln(e·e) = ln(e²) = 2, but (ln e)(ln e) = 1·1 = 1. Multiplication inside becomes addition outside, not multiplication.
Can I split ln(a + b) into ln a + ln b?
Never. ln(2+3) = ln 5 ≈ 1.61, but ln 2 + ln 3 = ln 6 ≈ 1.79. The law splits products, not sums — this is the most expensive mistake in the chapter.
Why does the product law work?
It is the mirror image of ex+y = ex·ey. Since ex turns addition into multiplication, its inverse ln must turn multiplication back into addition.
Should I split logs before differentiating?
Yes — d/dx [ln(2x)] becomes d/dx [ln 2 + ln x] = 1/x, with the constant vanishing. Splitting first dodges the chain rule entirely.
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