A logarithm asks 'what power?': Write logₐ x = y to mean aʸ = x: "to what power do I raise the base a to get x?"
Multiplying numbers adds their powers, so logarithms turn multiplication into addition — that is the engine behind all three laws and the reason log scales (decibels, the Richter scale, star magnitude) tame huge numbers into a friendly range.
All three laws need the same base on every log.
Two handy special values: logₐ 1 = 0 (since a⁰ = 1) and logₐ a = 1 (since a¹ = a).
These tidy up the end of many calculations — for example log(x/x) = log 1 = 0.
IB-style question — combine into one logarithm
A sound engineer writes a level as L = 2 log P + log Q − log 5, where P and Q are powers.
Write L as a single logarithm.
Step by step
- Use the power law on 2 log P to pull the 2 up as an exponent.
- Now it is log + log − log. Add the first two with the product law.
- Subtract the last term with the quotient law.
Final answer
L = log(P²Q / 5). Power first, then product (add), then quotient (subtract).
IB-style question — split a logarithm
Given that log 2 = 0.301 and log 3 = 0.477, find log 12 without a calculator.
Step by step
- Factorise 12 into the numbers you know: 12 = 2² × 3.
- Product law splits the multiplication into a sum.
- Power law brings the 2 down in front.
- Substitute the given values.
Final answer
log 12 ≈ 1.079. Build the number from its factors, then read off each known log.
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Any base, on demand: Your GDC computes logs in base 10 (`log`) and base e (`ln`). To evaluate a log in a different base — say log₂ 50 for a doubling-time question — rewrite it as a ratio of logs you can compute.
Pick the new base freely; base 10 or base e both work and give the same answer.
IB-style question — doubling time of an investment
An account grows by 6% per year, so after t years a deposit is multiplied by 1.06ᵗ. The doubling time satisfies 1.06ᵗ = 2.
Find t.
Step by step
- Take logs of both sides — any base; here base 10.
- Power law pulls t out of the exponent. This is the key move for exponential equations.
- Divide to isolate t, then evaluate on the GDC.
Final answer
t ≈ 11.9 years. Logs turn the unknown exponent into a coefficient you can divide out.
IB-style question — solve a log equation
Solve log₂(x + 6) = 5 for the size x of a data buffer.
Step by step
- A log equation undoes by rewriting in index form: logₐ N = k means aᵏ = N.
- Evaluate the power and solve.
- Check x + 6 = 32 > 0, so the logarithm is defined — valid.
Final answer
x = 26. Convert log = number into basenumber = inside, then solve.
Always check the domain: You can only take the log of a positive number. After solving, put every answer back into the original expressions: if any log ends up taking a zero or negative inside, reject that solution — it is not valid even though the algebra produced it.