SL 1.9 LAWS OF LOGARITHMS

📌 Key Definitions

Term Meaning
logax The power to which the base a must be raised to obtain x.
If ak = x, then logax = k.
Base The fixed number a in logax. In this topic, a = 10 or a = e only.
Domain We must have a > 0, a ≠ 1, and x > 0. Logarithms of 0 or negatives are not real.
🌍 Real-World Connection: Logarithms appear in pH (chemistry), Richter scale (earthquakes),
decibels (sound), and in analysing exponential growth and decay in biology and economics.

📌 Understanding Logarithms

Logarithms are the inverse of exponentials.
If 103 = 1000, then log10(1000) = 3.
If e2 ≈ 7.389, then ln(7.389) = 2.

  • They answer the question: “What exponent produced this number?”
  • Using laws of logarithms lets us simplify expressions, solve equations, and transform data.
  • In AHL 1.9, we assume the base is 10 (log) or e (ln).
🧠 Examiner Tip: Always check that the argument of a log is positive.
If you get log(−2) or log(0), something has gone wrong in your working.

📌 Law 1 – Product Law

Statement (for a > 0, a ≠ 1, x > 0, y > 0):

loga(x·y) = logax + logay

Intuition:

  • Multiplying inside the log becomes adding exponents.
  • If ap = x and aq = y, then xy = ap·aq = ap+q.
  • Therefore loga(xy) is the single exponent p+q, i.e. logax + logay.

Worked Example 1: Simplify log10(2000).

  1. Write 2000 as 2 × 1000 = 2 × 103.
  2. log(2000) = log(2 × 103) = log(2) + log(103).
  3. log(103) = 3, so log(2000) = log(2) + 3.

Worked Example 2: Expand ln(5e).

  1. ln(5e) = ln(5 × e) = ln(5) + ln(e).
  2. ln(e) = 1, so ln(5e) = ln(5) + 1.

📌 Law 2 – Quotient Law

Statement (for a > 0, a ≠ 1, x > 0, y > 0):

loga(x ÷ y) = logax − logay

Intuition:

  • Division inside the log becomes subtracting exponents.
  • If ap = x and aq = y, then x ÷ y = ap−q.
  • So loga(x ÷ y) = p − q = logax − logay.

Worked Example: Simplify log(50) − log(2).

  1. Combine using the quotient law in reverse: log(50) − log(2) = log(50 ÷ 2).
  2. 50 ÷ 2 = 25, so log(50) − log(2) = log(25).

📌 Law 3 – Power Law

Statement (for a > 0, a ≠ 1, x > 0):

loga(xm) = m·logax

Intuition:

  • Raising x to a power multiplies the exponent.
  • If x = ak, then xm = akm.
  • So loga(xm) = km = m·logax.

Logarithms-1.png

Worked Example 1: Expand log(√x) (base 10).

  1. √x = x1/2, so log(√x) = log(x1/2).
  2. Using the power law: log(x1/2) = (1/2)·log(x).

Worked Example 2: Simplify ln(e−3).

  1. ln(e−3) = −3·ln(e).
  2. ln(e) = 1, so ln(e−3) = −3
📱 GDC Tip: Use your calculator to check simplified expressions numerically.
For example, confirm that log(24) and log(3) + log(8) give the same decimal value.
📐 IA Spotlight:
You can model exponential growth/decay data (e.g. cooling, bacteria growth, finance)
and then use logarithms to linearise the relationship, allowing you to fit a straight line and estimate parameters.🌐
📝 Paper 1 Strategy:
When simplifying an expression with logs:

  • First, rewrite products, quotients and powers using the three laws.
  • Then combine like terms (for example 2logx − logx = logx).
  • Keep the base consistent: do not mix log base 10 and ln unless the question explicitly changes base.