ZIMSEC A Level · Pure Mathematics · Form 5
Laws of logarithms
- logₐ(xy) = logₐx + logₐy; logₐ(x/y) = logₐx − logₐy; logₐ(xⁿ) = n logₐx.
- logₐa = 1; logₐ1 = 0; change of base: logₐx = log x / log a.
The exponential function
eˣ (e ≈ 2.718) and its inverse ln x (natural logarithm): e^(ln x) = x and ln(eˣ) = x.
Solving equations
Worked example: Solve 3ˣ = 20.
x ln 3 = ln 20 → x = ln 20 / ln 3 ≈ 2.727
Worked example: Solve 2e²ˣ − 7eˣ + 3 = 0.
Let u = eˣ: 2u² − 7u + 3 = 0 → (2u − 1)(u − 3) = 0 → u = ½ or 3
x = ln ½ = −ln 2 ≈ −0.693, or x = ln 3 ≈ 1.099
Modelling and linear forms
- Exponential growth/decay: N = N₀e^(kt).
- y = axⁿ becomes ln y = ln a + n ln x (plot ln y against ln x: gradient n).
- y = abˣ becomes ln y = ln a + x ln b (plot ln y against x).
Exam tip: Check solutions of log equations – the argument of a logarithm must be positive.
Check yourself
- Solve log₂(x + 3) = 4.
- Express 2 log 3 + log 4 − log 6 as a single logarithm.
- A population P = 500e^(0.04t). Find t when P = 1 000.
Show answers
- x + 3 = 16, so x = 13
- log(9 × 4/6) = log 6
- e^(0.04t) = 2 → t = ln 2/0.04 ≈ 17.3
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