Logs as the reverse of exponents, plus every exam type — evaluating, converting between log and exponent form, the log laws, change of base, and solving exponential equations with logs — each worked in full, with a worksheet.
A logarithm answers one question: 'what power do I raise the base to, to get this number?' It's the exact reverse of an exponent — that single idea makes logs easy.
Read log2 8 as 'the power that turns 2 into 8'. Since 23 = 8, log2 8 = 3.
1Type 1: Evaluate a logarithm
- 1Ask: 5 to what power gives 125?The meaning of the log.
- 253 = 125.5 × 5 × 5 = 125.
- 3So log₅ 125 = 3.The logarithm is the exponent.
2Type 2: Convert between log and exponent form
- 124 = 16 → log₂ 16 = 4.Base stays the base; the exponent becomes the log's value.
- 2log₃ 81 = 4 → 34 = 81.Reverse the same relationship.
3Type 3: The log laws
- 1Product law: log 4 + log 25 = log(4 × 25) = log 100.Adding logs → multiply the numbers.
- 2log₁₀ 100 = 2.102 = 100.
4Type 4: Change of base
Calculators only do base 10 ('log') and base e ('ln'). For any other base, use: logb y = log ylog b.
- 1log₂ 20 = log 20 ÷ log 2.Change-of-base rule (any base on the calculator).
- 2≈ 1,301 ÷ 0,301 ≈ 4,32.Use the log button.
5Type 5: Solve an exponential equation with logs
- 1Take log of both sides: log 3ˣ = log 40.Logs bring the exponent down.
- 2x · log 3 = log 40 → x = log 40 ÷ log 3.Power law: log 3ˣ = x log 3.
- 3x ≈ 1,602 ÷ 0,477 ≈ 3,36.
Logs let you solve for an unknown in the exponent — e.g. 'how long until an investment doubles?' Take the log of both sides to bring the power down where you can solve it. Logs are also the inverse of the exponential graph.
log(a + b) is NOT log a + log b. The laws work on products and quotients — log(ab) = log a + log b — never on sums. One of the most common log errors.
Logs appear in finance, functions and exponential equations across matric. Master these five types in the worksheet.
6The graph when the base is between 0 and 1
If b > 1 the log graph increases. If 0 < b < 1 it decreases — it is the first graph reflected in the x-axis. Either way the domain is x > 0, the range is y ∈ ℝ, and the y-axis is a vertical asymptote.
log½ 8 = −3, not 3. A base below 1 gives negative outputs for x > 1 — check the sign against the graph before you write it down.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Evaluate log₂ 16. (evaluate)
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24 = 16 → 4 - 2Evaluate log₃ 27. (evaluate)
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33 = 27 → 3 - 3Evaluate log₁₀ 1000. (evaluate)
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103 = 1000 → 3 - 4Write 53 = 125 in log form. (convert)
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log₅ 125 = 3 - 5Write log₂ 32 = 5 in exponent form. (convert)
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25 = 32 - 6Simplify log 2 + log 50. (laws)
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log(2×50) = log 100 = 2 - 7Simplify log 200 − log 2. (laws)
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log(2002) = log 100 = 2 - 8Write log 8 as a multiple of log 2. (laws)
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log 23 = 3 log 2 - 9Evaluate log₂ 10 (change of base, 2 dp). (change base)
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log10 ÷ log2 ≈ 3,32 - 10Solve 2ˣ = 20 (2 dp). (solve)
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x = log20 ÷ log2 ≈ 4,32
Now practise it
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