Log laws

Combine, split, or use given log values

In 6 of 20 recent exams
Paper 1, no calculator
1 to 5 marks

How it is set in the exam

Paper 1 (no calculator) only, 1–5 marks per part: a short Section A question of its own, or the middle parts of a log question. Common (6 of 20 exam versions, 12 parts, all Paper 1 (no calculator)).

Spot it in the exam

Exam phrases to spot (4)

“write as a single logarithm”
“express in the form $\ln k$ ”
“given that $\ln 2 = a$ and $\ln 7 = b$ , express … in terms of $a$ and $b$ ”
“show that $\log_{a^2}(\ldots) = \ldots$ ”

In disguise (4)

a long sum of logs whose terms cancel in pairs (a telescoping sum), written with "…" or in sigma notation (a $\sum$ sign alone does not make it Sigma notation: find the value, or find the unknown: no arithmetic or geometric formula applies)
a "show that" linking two different bases
"find the value of …" when the values of other logs are given: the given logs have to be used, so it is this type, not an evaluation
"hence write $f(x)$ in the form $a + b\log x$ ": the rewrite is this type's move, but when the question then asks for the transformation it is Name the transformation

Variants (5)

combine: a sum, a multiple, or minus the log of a fraction, each into $\ln k$
split and substitute: the log of a composite number, in letters for the logs of its primes; the same in another base, by change of base
given log values: a log to a product base, $\log_{cd} c$ , is given; it gives $\log_{cd} d$ , then the logs of products, powers and roots of $c$ and $d$
show that: a log in base $b^2$ (or $b^3$ ) of a perfect square (or cube), as a log or a sum of logs in base $b$
a telescoping sum to one log, then the same sum with $n$ terms

Looks like it, but isn't (7)

the logs of the terms of a sequence ( $\ln x + \ln x^2 + \ln x^3 + \cdots$ is arithmetic in $\ln x$ ): Logs of a sequence
the unknown is inside a log or is its base: Solve a logarithmic equation, which uses these laws first
a relation between two unknowns, shown from a log equation: the first part of Solve a logarithmic equation
a constant $k$ in a model: Exponential and log models
one log of a number to evaluate, with no log values given and nothing to combine: Exact value of a logarithm
"describe the transformations" after the rewrite: Name the transformation
an exact area or integral in the form $p + q\ln r$ : Integrate: choose the method, Area between a curve and the x-axis or Area between two curves; only its last line uses these laws (Where exponents and logs hide)

Worked example

Write as ln k: the fractional and negative traps

Paper 1 (no calculator)

Question

Express each as a single natural logarithm, $\ln k$ , with $k \in \mathbb{Q}$ .
(a) $\ln 20 - \ln 4$ [1]
(b) $\frac{1}{2}\ln 81 + \ln 2$ [2]
(c) $-2\ln \frac{2}{3}$ [2]

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Which method?

The task	Move
Combine into one log	power law first (multipliers go inside as powers), then product and quotient; a constant becomes a log
Split and substitute	prime-factorise the argument; product, quotient and power laws; replace each log of a prime by its letter
Different bases	change of base to the base of the given logs (or to the prime), then the power law
Given the log of one factor to a product base ( $\log_{cd} c$ )	$\log_{cd}(cd) = 1$ gives $\log_{cd} d = 1 - \log_{cd} c$ ; then split the target into logs of $c$ and $d$
Telescoping	each term as the log of one fraction; the product law makes one log of a product; cancel; if nothing cancels at once, factorise each numerator and denominator first ( $k^2 - 1 = (k - 1)(k + 1)$ )

Steps

Power law first:
multipliers become powers (to combine), or powers come down (to split). Roots are powers too.
Product and quotient laws:
combine into one log, or split into logs of primes.
Change of base
where the bases differ: rewrite in the base the question uses.
Finish:
work out the power (the final mark needs $\ln 125$ , not $\ln 5^3$ ), or substitute the given letters or values.
"Show that":
start from one side, give each law its own line, and end on exactly the given form (the LHS-to-RHS layout, "show that" questions).

Common mistakes

Multiplying the multiplier into the argument
$\frac{1}{2}\ln 81$ is not $\ln 40.5$ . It is $\ln 81^{\frac{1}{2}} = \ln 9$ .
Splitting the log of a sum
$\ln(3 + 5) \ne \ln 3 + \ln 5$ . Only $\ln(3 \times 5)$ splits.
Leaving the power unworked
" $\ln 5^3$ " for the form $\ln k$ loses the final mark; write $\ln 125$ .

More mistakes (8)

A negative multiplier
$-\ln \frac{1}{3} = \ln 3$ , and $-2\ln\frac{2}{3} = \ln \frac{9}{4}$ : the fraction flips.
Cancelling letters in a quotient of logs
$\frac{2a}{a + b}$ is the final form. And $\frac{\ln 9}{\ln 15}$ is not $\ln \frac{9}{15}$ .
Change of base upside down
$\log_{15} 9 = \frac{\ln 9}{\ln 15}$ : the argument goes on top, the base underneath. Check: $9 < 15$ , so $\log_{15} 9 < 1$ , and $\frac{2a}{a + b} < 1$ .
A "show that" with jumps
each law carries its own method mark, and the last mark is only for working that leads straight to the given result, so the given answer written after a jump loses that last mark, and a law that is never shown loses its method mark. Give each law its own visible line.
A bare answer when log values are given
a markscheme can cap it. In one 4-mark part that used given log values, a correct value with no working scored only the first and last marks (2 of 4). Show the split and the substitution.
Letters with no line before them
write the line in the logs of the primes ( $\ln 3 + 2\ln 5$ ) before substituting $a$ and $b$ . That line has its own mark, so it still scores if the final expression goes wrong.
Treating a telescoping sum as a sequence
its terms are not in arithmetic or geometric sequence, so no $S_n$ formula applies.
A term left as "1 plus or minus a fraction" in a telescoping sum
write each term as the log of one fraction first, e.g. $\log_3\left(1 + \frac{2}{2k - 1}\right) = \log_3 \frac{2k + 1}{2k - 1}$ (the terms of Example 2.3). That line has its own mark, and nothing cancels until it is done.

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