Papers 2 and 3. \(\log_a x\) answers one question: what power of \(a\) gives \(x\)? Switch between log and index form, use the three laws to combine and split logarithms, and solve equations involving logs.
\(\log_a x = y \iff a^y = x\) (for \(a > 0\), \(a \ne 1\), \(x > 0\)). The laws: \(\log_a xy = \log_a x + \log_a y\), \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\), \(\log_a x^k = k\log_a x\). Always check that every logarithm in the original equation is of a positive number.
Fig. 1: y = log2 x is the reflection of y = 2^x in the line y = x. Each point (p, q) on one graph matches (q, p) on the other, e.g. (1, 2) and (2, 1)
Section 01
Logs and indices are the same fact.
\(2^5 = 32\) and \(\log_2 32 = 5\) say the same thing: the power of 2 that gives 32 is 5. Every log question can be turned into an index question and back. Because \(a^y\) is always positive, you can only take the log of a positive number.
Definition: \(\log_a x = y \iff a^y = x\), with base \(a > 0\), \(a \ne 1\).
Special values: \(\log_a a = 1\), \(\log_a 1 = 0\), \(\log_a a^k = k\), and \(a^{\log_a x} = x\).
Base 10: \(\log_{10} x\) is often written \(\lg x\). Base \(e\) (\(\ln x\)) comes in the next lesson.
Graph: \(y = \log_a x\) is the reflection of \(y = a^x\) in \(y = x\) (Fig. 1). It passes through \((1, 0)\) and exists only for \(x > 0\).
Worked example Switching forms
(a) Write \(\log_3 81 = 4\) in index form. (b) Write \(2^{-3} = \dfrac{1}{8}\) in log form.
(a) The base 3 raised to the answer 4 gives 81.
(b) The base is 2, the power is \(-3\), the result is \(\dfrac{1}{8}\).
Find the exact value of (a) \(\log_5 \sqrt{5}\) (b) \(\log_4 8\).
(a) \(\sqrt{5} = 5^{1/2}\), so \(\log_5 5^{1/2} = \dfrac{1}{2}\).
(b) Let \(\log_4 8 = y\). Then \(4^y = 8\), i.e. \(2^{2y} = 2^3\).
So \(2y = 3\) and \(y = \dfrac{3}{2}\).
Answer: (a) \(\dfrac{1}{2}\) (b) \(\dfrac{3}{2}\)
Examiner trap
\(\log_2 \dfrac{1}{8}\) is \(-3\), not \(\dfrac{1}{3}\) or \(-\dfrac{1}{3}\). Negative logs come from fractions; fractional logs come from roots.
Section 02
The three laws.
The laws of logarithms are the index laws in disguise: multiplying powers adds the indices, so the log of a product is the sum of the logs. Use them to combine several logs into one, or to split one log into simpler parts.
Index law
Log law
\(a^m \times a^n = a^{m + n}\)
\(\log_a xy = \log_a x + \log_a y\)
\(a^m \div a^n = a^{m - n}\)
\(\log_a \dfrac{x}{y} = \log_a x - \log_a y\)
\((a^m)^k = a^{mk}\)
\(\log_a x^k = k \log_a x\)
Worked example Combining into a single log
Express \(2\log_a 3 + \log_a 4 - \log_a 6\) as a single logarithm.
Power law first: \(2\log_a 3 = \log_a 3^2 = \log_a 9\).
There is no law for \(\log(x + y)\): \(\log_a(x + y) \ne \log_a x + \log_a y\). Also \(\dfrac{\log_a x}{\log_a y}\) is NOT \(\log_a \dfrac{x}{y}\).
Stage 1 · Learn
Check your understanding
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What is \(\log_2 32\)?
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Which is \(10^{-2} = 0.01\) written in log form?
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\(\log_a 5 + \log_a 7\) equals
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For any valid base \(a\), what is \(\log_a 1\)?
Section 03
Solving log equations.
Collect the logs on one side into a single log using the laws, then convert to index form. Finally, check every answer in the ORIGINAL equation: any value that makes the inside of a log zero or negative must be rejected.
Check: both \(x = 3\) and \(x = 6\) are greater than 2, so every log is of a positive number.
Answer: \(x = 3\) or \(x = 6\)
Examiner trap
Do not 'cancel the logs' term by term: \(\log_2(x + 3) + \log_2(x - 1) = 5\) does NOT give \((x + 3) + (x - 1) = 5\) or \(= 32\). Make ONE log first, then undo it.
Stage 2 · Exam
Exam-style questions
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Without a calculator, find the value of \(2\log_3 6 - \log_3 4\).
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Solve \(\log_2(x + 3) + \log_2(x - 1) = 5\).
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Given that \(\log_a x = p\) and \(\log_a y = q\), express \(\log_a \dfrac{a^2 x}{\sqrt{y}}\) in terms of \(p\) and \(q\).
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Solve \(\log_3(2x + 1) - \log_3(x - 2) = 2\).
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