Home / IGCSE Add Maths 0606 / Logarithmic and exponential functions / Laws of logarithms and change of base
Cambridge IGCSE · Topic 6.2
IGCSE Add Maths 0606 / Topic 6 / IGCSE

Logs turn products into sums.

A logarithm is a power, so the rules for powers become rules for logs: multiplying adds logs, dividing subtracts them and a power comes down to the front. With these laws, and the change of base rule, you can combine several logs into one, split one into several and find exact values like \(\log_{8} 32\) without a calculator.

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The key idea

For \(x, y > 0\): \(\log_a (xy) = \log_a x + \log_a y\), \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\) and \(\log_a x^{k} = k\log_a x\). Also \(\log_a a = 1\) and \(\log_a 1 = 0\); \(\lg\) means \(\log_{10}\) and \(\ln\) means \(\log_{e}\). Change of base: \(\log_a b = \dfrac{\log_c b}{\log_c a}\), so in particular \(\log_a b = \dfrac{1}{\log_b a}\). There is NO law for \(\log_a (x + y)\).

x y O 1 2 4 8 1 2 3 −1 −2 (4, 2) (4, 1) (8, 3) (8, 1.5) (1, 0) y = log₂ x y = log₄ x Change of base: log₄ x = log₂ x ÷ log₂ 4 = ½ log₂ x Every height on the green curve is half the blue one. Both pass through (1, 0): log 1 = 0
Fig. 1: the graphs of y = log2 x (blue) and y = log4 x (green). By the change of base rule, log4 x = log2 x / log2 4 = (1/2) log2 x, so each height on the green curve is half the height on the blue curve: at x = 4 the heights are 2 and 1, at x = 8 they are 3 and 1.5. Both curves pass through (1, 0) because the log of 1 is 0 in any base
Section 01

Three laws turn products into sums.

Because \(\log_a x\) is the power that \(a\) must be raised to to give \(x\), the index laws give three log laws. Multiplying numbers adds their powers, so \(\log_a (xy) = \log_a x + \log_a y\). Dividing subtracts them, so \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\). Raising to a power multiplies the power, so \(\log_a x^{k} = k\log_a x\). Two special values come straight from the definition: \(a^{1} = a\) gives \(\log_a a = 1\), and \(a^{0} = 1\) gives \(\log_a 1 = 0\). Using the laws from left to right EXPANDS a log into simpler pieces.

Worked example Expanding a single log
Express \(\log_3 \dfrac{9x^{2}}{y}\) in terms of \(\log_3 x\) and \(\log_3 y\).
  1. Quotient law: \(\log_3 \dfrac{9x^{2}}{y} = \log_3 (9x^{2}) - \log_3 y\).
  2. Product law: \(\log_3 (9x^{2}) = \log_3 9 + \log_3 x^{2}\).
  3. \(\log_3 9 = 2\) because \(3^{2} = 9\), and the power law gives \(\log_3 x^{2} = 2\log_3 x\).
  4. Put it together: \(2 + 2\log_3 x - \log_3 y\).
Answer: \(2 + 2\log_3 x - \log_3 y\)
Worked example Exact value from two logs
Without using a calculator, find the value of \(\log_6 4 + \log_6 9\).
  1. Neither log is a whole number on its own, so combine them first.
  2. Product law: \(\log_6 4 + \log_6 9 = \log_6 (4 \times 9) = \log_6 36\).
  3. \(6^{2} = 36\), so \(\log_6 36 = 2\).
Answer: \(2\)
Your turn
Without a calculator, find \(\log_2 40 - \log_2 5\).
Examiner trap

There is no law for a sum inside a log: \(\log (x + y)\) is NOT \(\log x + \log y\) (that equals \(\log (xy)\)). Also \((\log x)^{2}\) is NOT \(2\log x\): the power law only brings down a power of \(x\) itself, \(\log x^{2} = 2\log x\). And \(\dfrac{\log x}{\log y}\) is NOT \(\log x - \log y\); the quotient law is about \(\log \dfrac{x}{y}\).

Section 02

Write it as a single logarithm.

Using the laws from right to left COMBINES several logs into one. Do it in a fixed order: first move every number in front of a log up as a power (\(3\lg a = \lg a^{3}\)), then turn any plain number into a log in the same base, then combine: logs with a plus sign go on top, logs with a minus sign go underneath. The key fact for the constants is \(\log_a a^{k} = k\): in base 10, \(1 = \lg 10\), \(2 = \lg 100\) and \(3 = \lg 1000\); in base \(e\), \(1 = \ln e\).

Worked example A constant and two lg terms
Write \(2 + 3\lg a - \lg b\) as a single logarithm to base 10.
  1. Power law: \(3\lg a = \lg a^{3}\).
  2. Constant: \(2 = \lg 10^{2} = \lg 100\).
  3. Combine: \(\lg 100 + \lg a^{3} - \lg b = \lg \dfrac{100a^{3}}{b}\).
  4. Check: expanding \(\lg \dfrac{100a^{3}}{b}\) gives \(2 + 3\lg a - \lg b\).
Answer: \(\lg \dfrac{100a^{3}}{b}\)
Worked example Natural logs and a 1
Write \(1 + 2\ln x - \ln (x + 1)\) as a single natural logarithm.
  1. Power law: \(2\ln x = \ln x^{2}\).
  2. Constant: \(1 = \ln e\).
  3. Combine: \(\ln e + \ln x^{2} - \ln (x + 1) = \ln \dfrac{e x^{2}}{x + 1}\).
  4. Note that \(\ln (x + 1)\) stays as one piece: it cannot be split into \(\ln x + \ln 1\).
Answer: \(\ln \dfrac{e x^{2}}{x + 1}\)
Your turn
A preview of the next lesson: solve \(\lg x + \lg (x - 3) = 1\).
Examiner trap

Move the coefficients up BEFORE combining: \(2 + 3\lg a - \lg b\) is not \(\lg \dfrac{2 \times 3a}{b}\). A plain number must become a log first: \(2 + \lg x\) is \(\lg (100x)\), not \(\lg (2x)\) or \(\lg (x + 2)\). And a minus sign applies to the whole log after it: \(\lg p - 2\lg q = \lg \dfrac{p}{q^{2}}\), not \(\lg (p q^{2})\).

Stage 1 · Learn

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Without a calculator, find \(\log_2 24 - \log_2 3\).

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Which of these is equal to \(\log_5 x^{3}\)?

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Write \(1 + \lg 3\) as a single logarithm to base 10.

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Find the exact value of \(\log_9 27\).

Section 03

Change of base.

Your calculator has only \(\lg\) and \(\ln\), and many exact values are easier in a different base, so you need to switch bases. If \(y = \log_a b\) then \(a^{y} = b\); taking \(\log_c\) of both sides gives \(y\log_c a = \log_c b\), so \(\log_a b = \dfrac{\log_c b}{\log_c a}\). The new base \(c\) can be anything: choose one in which both logs are easy. Putting \(c = b\) gives the special case \(\log_a b = \dfrac{1}{\log_b a}\): swapping the base and the number turns the log upside down. Fig. 1 shows the rule as a picture: \(\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \tfrac{1}{2}\log_2 x\).

Worked example Exact values without a calculator
Find the exact value of (a) \(\log_8 32\) (b) \(\log_4 \dfrac{1}{8}\).
  1. (a) 8 and 32 are both powers of 2, so change to base 2: \(\log_8 32 = \dfrac{\log_2 32}{\log_2 8}\).
  2. \(\log_2 32 = 5\) and \(\log_2 8 = 3\), so \(\log_8 32 = \dfrac{5}{3}\). (Check: \(8^{5/3} = (\sqrt[3]{8})^{5} = 2^{5} = 32\).)
  3. (b) Again use base 2: \(\log_4 \dfrac{1}{8} = \dfrac{\log_2 2^{-3}}{\log_2 2^{2}} = \dfrac{-3}{2}\).
  4. Check: \(4^{-3/2} = \dfrac{1}{(\sqrt{4})^{3}} = \dfrac{1}{8}\).
Answer: (a) \(\frac{5}{3}\) (b) \(-\frac{3}{2}\)
Worked example Turning a reciprocal into a natural log
Write \(\dfrac{3}{\log_2 e}\) as a single natural logarithm.
  1. Swap rule: \(\dfrac{1}{\log_2 e} = \log_e 2 = \ln 2\).
  2. So \(\dfrac{3}{\log_2 e} = 3\ln 2\).
  3. Power law: \(3\ln 2 = \ln 2^{3} = \ln 8\).
Answer: \(\ln 8\)
Your turn
Given that \(\log_3 x = p\), express \(\log_9 x\) and \(\log_x 27\) in terms of \(p\).
Examiner trap

\(\dfrac{\log_c b}{\log_c a}\) is a DIVISION of two logs: it is not \(\log_c \dfrac{b}{a}\) and not \(\log_c b - \log_c a\). Keep the old base on the bottom: \(\log_8 32 = \dfrac{\log_2 32}{\log_2 8} = \dfrac{5}{3}\), not \(\dfrac{3}{5}\). And \(\log_9 x\) is \(\tfrac{1}{2}\log_3 x\), not \(2\log_3 x\): a bigger base gives a SMALLER log, as Fig. 1 shows.

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Without using a calculator, find the value of \(2\lg 5 + \lg 12 - \lg 3\).

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Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express \(\log_a (18a)\) in terms of \(p\) and \(q\).

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Given that \(\log_2 x = t\), express \(\log_x 8 + \log_4 x\) in terms of \(t\).

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Write \(\dfrac{4}{\log_3 e}\) as a single natural logarithm.

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Go deeper · practice
Six original Cambridge-style questions
Exact values from the three laws, a single base 10 log with a constant, logs in terms of \(p\) and \(q\), change of base without a calculator, a reciprocal-log identity and a three-part change-of-base question.