The Laws of Logarithms

Use the product, quotient and power laws and change of base to combine logs into one, find exact values and rewrite logs in other bases.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 13 real past-paper questions below

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What a log asks

Before calculators, scientists had to multiply huge numbers by hand. Logarithms turned that slow multiplying into quick adding. The laws in this lesson are the reason it works.

Remember what a log asks: what power do I need? For example, $lo g_{2} 8 = 3$ because $2^{3} = 8$ .

Rule

$lo g_{a} x = y$ means exactly the same as $a^{y} = x$ . The log is the power.

Two facts come straight from this. Any number to the power 1 is itself, and any number to the power 0 is 1.

Formula

lo g_{a} a = 1

$a$ — the base. It can be any positive number except 1. When the base and the number inside match, the log is 1.

Formula

lo g_{a} 1 = 0

$1$ — the number inside. The log of 1 is 0 in every base, because $a^{0} = 1$ .

Worked example

1
Find the value of $lo g_{5} 5 + l g 1 + lo g_{2} 8$ without a calculator.
2
$lo g_{5} 5 = 1$ — Base and number match.
3
$l g 1 = 0$ — The log of 1 is always 0.
4
$lo g_{2} 8 = 3$ — 2 to the power 3 is 8.
5
$1 + 0 + 3 = 4$ — Add the three values.

Notation: $l g x$ means $lo g_{10} x$ , and $ln x$ means $lo g_{e} x$ . Every law in this lesson works for every base.

Rule

Learn all the laws in this lesson. 0606 and 4037 print none of them. 4PM1 prints only the change-of-base formula on the formula sheet.

Key idea

John Napier, a Scottish mathematician, published the first table of logarithms in 1614. For more than 300 years, astronomers and ship navigators multiplied big numbers by looking up logs and adding them.

Product and quotient laws

When you multiply powers of the same base, you add the indices: $a^{m} \times a^{n} = a^{m + n}$ . A log is an index, so logs follow the same pattern.

Formula

lo g_{a} (x y) = lo g_{a} x + lo g_{a} y

$x$ and $y$ — any two positive numbers, multiplied inside one log.
$a$ — the base. It must be the same in all three logs.

Dividing powers means subtracting indices, $a^{m} \div a^{n} = a^{m - n}$ . So a division inside a log becomes a minus outside.

Formula

lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y

$x$ — the number on top. Its log is the one you keep.
$y$ — the number underneath. Its log is the one you subtract.

Worked example

1
Find the exact value of $lo g_{6} 4 + lo g_{6} 9$ .
2
$lo g_{6} 4 + lo g_{6} 9 = lo g_{6} (4 \times 9)$ — Same base and a plus sign: use the product law.
3
$= lo g_{6} 36$ — 4 × 9 = 36.
4
$= 2$ — 6 squared is 36, so the power is 2.

Worked example

1
Find the exact value of $lo g_{2} 40 - lo g_{2} 5$ .
2
$lo g_{2} 40 - lo g_{2} 5 = lo g_{2} (\frac{40}{5})$ — Same base and a minus sign: use the quotient law.
3
$= lo g_{2} 8$ — 40 ÷ 5 = 8.
4
$= 3$ — 2 cubed is 8.

Common mistake

$lo g_{a} (x + y)$ is NOT $lo g_{a} x + lo g_{a} y$ . The product law needs x times y inside. A sum inside a log cannot be split.

Common mistake

$\frac{l o g _{a} x}{l o g _{a} y}$ is NOT $lo g_{a} (\frac{x}{y})$ . The quotient law needs a minus sign between two logs, not a fraction line.

Key idea

Times inside becomes plus outside. Divide inside becomes minus outside. The base must be the same in every log.

The power law

A power of a power multiplies the indices: $(a^{m})^{k} = a^{mk}$ . So a power inside a log can move out to the front.

Formula

lo g_{a} (x^{k}) = k lo g_{a} x

$k$ — the power on x. It can be a whole number, a negative number or a fraction.
$x$ — the positive number inside the log.

Put $k = - 1$ and you get a special case that saves time:

Formula

lo g_{a} (\frac{1}{x}) = - lo g_{a} x

$x$ — the number under the 1. Flipping a number upside down just changes the sign of its log.

Worked example

1
Find the exact value of $lo g_{2} (\frac{1}{8})$ .
2
$\frac{1}{8} = 8^{- \frac{1}{2}}$ — A square root is the power one half. Being underneath makes the power negative.
3
$lo g_{2} 8^{- \frac{1}{2}} = - \frac{1}{2} lo g_{2} 8$ — Power law: the power moves to the front.
4
$= - \frac{1}{2} \times 3$ — 2 cubed is 8, so log base 2 of 8 is 3.
5
$= - \frac{3}{2}$ — Multiply.

Common mistake

$(lo g_{a} x)^{2}$ is NOT $2 lo g_{a} x$ . The power law only works when the power is on the x inside: $lo g_{a} x^{2} = 2 lo g_{a} x$ .

Rule

No calculator on 0606 and 4037 Paper 1: give exact answers such as $- \frac{3}{2}$ , never a rounded decimal.

Key idea

A power inside the log becomes a number in front. This is the law that brings an unknown power down to the ground.

Write as a single log

A favourite exam question: write a mix of logs and plain numbers as one single log. First you need one more fact: any plain number can be written as a log.

Formula

k = lo g_{a} (a^{k})

$k$ — any plain number in the expression.
$a$ — the base the question asks for. For example, $2 = l g 100$ and $1 = ln e$ .

Now work in the same order every time:

Steps

1
1. Power law: move every number in front of a log back inside as a power.
2. Turn every plain number into a log of the same base.
3. Combine: plus terms go on top, minus terms go underneath.

Worked example

1
Write $3 l g x - 2 l g y + 1$ as a single logarithm to base 10.
2
$3 l g x = l g x^{3}, 2 l g y = l g y^{2}$ — Power law first: each front number goes back inside as a power.
3
$1 = l g 10$ — Base 10, and 10 to the power 1 is 10.
4
$l g x^{3} - l g y^{2} + l g 10$ — Now every term is one single log.
5
$= l g (\frac{10 x ^{3}}{y ^{2}})$ — Plus terms on top, the minus term underneath.

Worked example

1
Write $2 ln a - ln (a + 4) - 3$ as a single logarithm to base e.
2
$2 ln a = ln a^{2}$ — Power law.
3
$3 = ln e^{3}$ — Base e, so 3 is the log of e cubed.
4
$ln a^{2} - ln (a + 4) - ln e^{3}$ — Every term is now a single ln.
5
$= ln (\frac{a ^{2}}{e ^{3} ( a + 4 )})$ — The plus term on top, both minus terms underneath.

Examiners give a mark for using the power law, then the rest for the correct single log. So show the power-law line.

Common mistake

Writing 2 as $l g 2$ . A plain number k becomes the log of base to the power k, so $2 = l g 100$ .

Common mistake

Splitting $ln (a + 4)$ into $ln a + ln 4$ . A sum inside a log stays together in its brackets.

Key idea

Powers in first, numbers into logs second, combine last. Same order, every question.

Build from known logs

Another common question gives you two logs as letters, then asks for a different log. The trick is to build the new number out of the numbers you know.

Worked example

1
Given that $l g 2 = p$ and $l g 3 = q$ , write $l g 72$ in terms of p and q.
2
$72 = 2^{3} \times 3^{2}$ — Write 72 using only 2s and 3s: 8 × 9.
3
$l g 72 = l g 2^{3} + l g 3^{2}$ — Product law: times inside becomes plus.
4
$= 3 l g 2 + 2 l g 3$ — Power law: the powers come to the front.
5
$= 3 p + 2 q$ — Swap in the letters.

Worked example

1
With the same p and q, write $l g 0.6$ in terms of p and q.
2
$0.6 = \frac{2 \times 3}{10}$ — 0.6 is 6 over 10, and 6 is 2 × 3.
3
$l g 0.6 = l g 2 + l g 3 - l g 10$ — Product law on top, quotient law for the 10.
4
$l g 10 = 1$ — Base and number match.
5
$l g 0.6 = p + q - 1$ — Swap in the letters.

Common mistake

Writing $l g 6 = l g 2 \times l g 3$ . Times inside becomes PLUS outside: $l g 6 = l g 2 + l g 3$ .

Key idea

Break the number into the pieces you know. Times becomes plus, divide becomes minus, powers come to the front.

Change of base

Exam questions often mix bases, like base 2 and base 4 in one line. You can only combine logs with the same base, so you need a way to change the base.

Here is why it works. Each line follows from the one before:

$lo g_{a} x = y$ — call the log y.

$a^{y} = x$ — what a log means.

$lo g_{b} (a^{y}) = lo g_{b} x$ — take a log of base b on both sides.

$y lo g_{b} a = lo g_{b} x$ — power law.

$y = \frac{l o g _{b} x}{l o g _{b} a}$ — divide both sides.

Formula

lo g_{a} x = \frac{lo g _{b} x}{lo g _{b} a}

$a$ — the old base, the one you want to get rid of.
$b$ — the new base. You choose it.
$x$ — the number inside the log.

Rule

4PM1 prints this formula on the formula sheet. 0606 and 4037 do not, so learn it.

Choose the new base to be b itself, and you get a quick swap rule. Learn it for every board.

Formula

lo g_{a} b = \frac{1}{lo g _{b} a}

$a$ and $b$ — the base and the number simply swap places, and you take 1 over the result.

Worked example

1
Write $lo g_{4} x$ in terms of $lo g_{2} x$ .
2
$lo g_{4} x = \frac{l o g _{2} x}{l o g _{2} 4}$ — Change to base 2, the base we want.
3
$lo g_{2} 4 = 2$ — 2 squared is 4.
4
$lo g_{4} x = \frac{1}{2} lo g_{2} x$ — Divide by 2.

Worked example

1
Write $\frac{1}{l o g _{2} e} + ln 3$ as a single natural logarithm.
2
$\frac{1}{l o g _{2} e} = lo g_{e} 2$ — Swap rule: the base and the number change places.
3
$lo g_{e} 2 = ln 2$ — Log base e is ln.
4
$ln 2 + ln 3 = ln 6$ — Product law: 2 × 3 = 6.

Worked example

1
Use a calculator to find $lo g_{7} 20$ to 3 significant figures.
2
$lo g_{7} 20 = \frac{l g 20}{l g 7}$ — Your calculator has lg, so change to base 10.
3
$\approx 1.54$ — 1.5395... rounded to 3 significant figures.

Common mistake

$\frac{l o g _{b} x}{l o g _{b} a}$ is NOT $lo g_{b} (\frac{x}{a})$ . A fraction of two logs is not the log of a fraction.

Key idea

Pick the base that the rest of the question uses, then change every log to that base.

Which law? Read the question

Before you start, read what the question asks. The wording tells you which law to reach for.

You see	You do
“Write as a single logarithm”	Power law first, plain numbers into logs, then combine
“Find the exact value”	Write the number inside as a power of the base
Given $l g 2 = p$ , “in terms of p”	Break the number into the pieces you know
Two different bases	Change of base to one base
$\frac{1}{l o g _{a} b}$	Swap: $lo g_{b} a$
A plain number next to logs	Write it as a log: $k = lo g_{a} a^{k}$

Common mistake

Using a law on logs with different bases. Change the bases to match first, then combine.

Key idea

Every law comes from the rules of indices. If you forget one, rebuild it from $a^{m} \times a^{n} = a^{m + n}$ .

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