What you need to know
Laws of exponents describe how powers combine. In each law the base must be the same: am × an = am + n, am ÷ an = am − n, (am)n = amn, (ab)n = anbn, a0 = 1, a−n = 1an, and a1/n = ⁿ√a, so am/n = (ⁿ√a)m.
A logarithm is the answer to the question "what power?". logₐ x = y means a[sup y] = x. For example log₂ 8 = 3 because 23 = 8. The base a must be positive and not equal to 1, and you can only take the log of a positive number. Two bases get special names: log₁₀ x is often written log x, and logₑ x (base e ≈ 2.718) is the natural logarithm, written ln x.
| Logarithm law | Exponent law it comes from |
|---|
| logₐ (xy) = logₐ x + logₐ y | am × an = am + n |
| logₐ (xy) = logₐ x − logₐ y | am ÷ an = am − n |
| logₐ (xn) = n logₐ x | (am)n = amn |
| logₐ a = 1 and logₐ 1 = 0 | a1 = a and a0 = 1 |
| logₐ x = ln xln a (change of base) | lets a calculator evaluate any base |
Simplify (2x3)4 ÷ 4x5
- (2x3)4 = 24 × x12 = 16x12
- 16x12 ÷ 4x5 = 4x12 − 5 = 4x[sup 7]
Evaluate 82/3 and 27−1/3 without a calculator
- 82/3 = (³√8)2 = 22 = 4
- 27−1/3 = 1271/3 = 1³√27 = [frac 1/3]
Evaluate log₂ 32, log₅ (125) and log₁₀ 0.001
- log₂ 32: 25 = 32, so log₂ 32 = 5
- log₅ (125): 5−2 = 125, so log₅ (125) = −2
- log₁₀ 0.001: 10−3 = 0.001, so log₁₀ 0.001 = −3
Simplify log₃ 54 − log₃ 2, and expand ln (x2√y)
- log₃ 54 − log₃ 2 = log₃ (542) = log₃ 27 = 3 (because 33 = 27)
- ln (x2√y) = ln x2 + ln y1/2 = 2 ln x + [frac 1/2] ln y
Solve 3x = 20, correct to 2 decimal places
- Take ln of both sides: ln 3x = ln 20
- Bring the power down: x ln 3 = ln 20
- x = ln 20ln 3 = 2.9957…1.0986… = 2.73
- Check: 32.73 ≈ 20.1, which is close to 20.
Solve log₂ x + log₂ (x − 2) = 3
- Combine into one log: log₂ [x(x − 2)] = 3
- Rewrite in index form: x(x − 2) = 23 = 8
- x2 − 2x − 8 = 0, so (x − 4)(x + 2) = 0, giving x = 4 or x = −2.
- Check the domain: log₂ x needs x > 0, so x = −2 is rejected. The only solution is x = 4.
Common mistake: there is no law for the log of a sum. log (x + y) is not log x + log y, and log x × log y is not log (xy). The laws only turn products, quotients and powers inside a log into sums, differences and multiples outside it. Another common mistake: forgetting to reject solutions that make the inside of a log zero or negative. Always substitute each solution back into the original equation before writing your final answer.
Check yourself
- Simplify x5 × x−2 ÷ x.
- Evaluate 163/4.
- Evaluate log₃ 81.
- Write log 2 + 2 log 5 as a single logarithm.
- Solve 2x = 10, correct to 2 decimal places.
- Solve log₅ (2x + 1) = 2.
Answers: 1. x5 − 2 − 1 = x2 2. (⁴√16)3 = 23 = 8 3. 4 (because 34 = 81) 4. log 2 + log 25 = log 50 5. x = ln 10ln 2 ≈ 3.32 6. 2x + 1 = 25, so x = 12