Each rule for logs mirrors a rule for exponents. Multiplying two powers of the same base adds their exponents, so adding logs means multiplying the inputs. Dividing subtracts them. Raising to a power multiplies them. Because a logarithm is an exponent in disguise, every exponent law has a direct log translation, and learning them in pairs makes both sets easier to remember.
Key Takeaways
- Product rule:
log10(100) + log10(10) = log10(1000) = 3, since 2 + 1.- Exponent twin:
2³ × 2⁴ = 2⁷ = 128, because multiplying powers adds exponents.- Power rule:
ln(e³) = 3 × ln(e) = 3.- Change of base:
log(8, 2) = 3, and any base works withlog(x) ÷ log(b).
The rules as a cheat sheet
The pairings below are the ones that do routine work. Each log rule on the right is the exponent rule on the left read backwards. The log and exponent rules guide keeps this same table, and exponents explained develops where the exponent rules come from.
| Exponent rule | Log rule | Quick check |
|---|---|---|
aⁿ × aᵐ = aⁿ⁺ᵐ |
log_b(xy) = log_b(x) + log_b(y) |
log10(100) + log10(10) = 2 + 1 = 3 |
aⁿ ÷ aᵐ = aⁿ⁻ᵐ |
log_b(x ÷ y) = log_b(x) − log_b(y) |
log10(1) − log10(100) = 0 − 2 = −2 |
(aⁿ)ᵐ = aⁿᵐ |
log_b(xⁿ) = n × log_b(x) |
ln(e³) = 3 × ln(e) = 3 |
a⁰ = 1 |
log_b(1) = 0 |
ln(1) = 0 |
a⁻ⁿ = 1 ÷ aⁿ |
log_b(1 ÷ x) = −log_b(x) |
ln(0.5) = −ln(2) = −0.693147 |
Every row above is reproducible on the calculator, with values rounded the way the display rounds them. The calculator function reference lists the exact keys.
Product rule: the log of a product is the sum of logs
This is the most useful log rule. log_b(xy) = log_b(x) + log_b(y). Multiplication, which is slow by hand, turns into addition, which is fast. The reason it works: if x = b^p and y = b^q, then xy = b^p × b^q = b^(p+q), so the log of xy is p + q.
Worked example 1. log10(100) = 2 and log10(10) = 1. The product rule says log10(100 × 10) = log10(1000) = 2 + 1 = 3. Check on the log calculator: type log(1000) and it shows 3. Better still, type log(100) + log(10) and it also settles on 3. Two paths, same answer.
The exponent twin is the one everyone meets first: 2³ × 2⁴ = 2⁷ = 128. Full working: 8 × 16 = 128, and 2⁷ = 128 as well.
Quotient rule: the log of a quotient is the difference
Division becomes subtraction: log_b(x ÷ y) = log_b(x) − log_b(y). If a product breaks apart into an addition, then a quotient breaks apart into a subtraction. The exponent version flips a fraction into a negative exponent first: 2⁵ ÷ 2² = 2³ = 8, because 5 minus 2 is 3.
Worked example 2. log10(1) − log10(100) should equal the log of 1 ÷ 100 = 0.01. That gives 0 − 2 = −2, and indeed log10(0.01) = −2 because 10⁻² = 0.01. Type log(0.01) to confirm. Notice the log of a number below 1 is negative, just as a negative exponent produces a fraction.
Power rule: the exponent comes out in front
log_b(xⁿ) = n × log_b(x). Any exponent inside the log becomes a multiplier in front. This is the rule that lets you solve for an exponent: log(5^10) becomes 10 × log(5).
Worked example 3. ln(e^3) uses the natural log calculator convention where ln is base e. The power rule turns it into 3 × ln(e), and since ln(e) = 1, the answer is exactly 3. Type ln(e^3) and 3 * ln(e); both return 3. On the exponent side, (2³)² = 2⁶ = 64.
Change of base: any log in any base
The rules give you a one division conversion between bases: log_b(x) = log(x) ÷ log(b). The log calculator also accepts a second argument for the base, so you can often skip the formula entirely.
Worked example 4. You want log2(8). On the log calculator, type log(8, 2) and it returns 3, because 2³ = 8. Or use the change of base formula: log(8) / log(2) also comes back 3. The exponent side shows the same relation in another costume: 2⁴ × 2⁻¹ = 2³ = 8.
Working with the rules on the calculator
- Type the whole expression, then press
=.log(100) + log(10)evaluates once and gives 3. - Any base works as a second argument:
log(x, b). - Remember that
logis base 10 here andlnis the natural version. The exponent calculator covers the same pair of ideas from the power side.
Frequently asked questions
Why does the product rule hold?
Because multiplying powers of the same base adds exponents. If x = b^p and y = b^q, then xy = b^(p+q), and taking a log of both sides gives log_b(xy) = p + q.
Do these rules work for natural logs?
Every rule works in every base, including e. ln(xy) = ln(x) + ln(y), ln(x ÷ y) = ln(x) − ln(y), and ln(xⁿ) = n × ln(x). The natural log calculator applies the same laws because a log is a log regardless of base.
What is the log of 1?
log_b(1) = 0 in every base, because any base raised to the power 0 equals 1.
How do I compute log2(8) if there is no base key?
Type log(8, 2) to set the base directly, or use the change of base formula log(8) / log(2). Both give 3.