You are not about to learn three new rules. You are about to watch three rules you already know get rewritten in logarithm form, and once you have seen the translation you will never need to memorise them.
Today's hook: You know that $2^3 \times 2^4 = 2^7$: multiplying powers adds the indices. A logarithm is an index. So what should happen to logarithms when you multiply the numbers they came from?
0/5QUESTS
Think First
warm-up
Write down the three index laws for $a^m \times a^n$, $a^m \div a^n$ and $(a^m)^n$. Beside each one, write what the index does: added, subtracted, or multiplied. Now predict: if a logarithm is an index, what will $\log_a (xy)$ turn out to equal? Commit to a guess before you read on.
Record your answer in your workbook.
1
The Big Idea
+5 XP to read
Every law of logarithms is an index law with the roles swapped. Multiplying two numbers adds their indices, so it adds their logarithms. Dividing subtracts them. Raising to a power multiplies. Three index laws in, three logarithm laws out, and the derivation is the same three lines each time.
$$\log_a(xy) = \log_a x + \log_a y$$
The proof always runs the same way: name the two logarithms, translate to index form, apply the index law you already know, then translate back. Learning that three-step shape is worth more than learning the three results, because it also reconstructs them when you forget.
$\log_a(xy) = \log_a x + \log_a y$
Same base throughout
Every law needs both logarithms to share a base. $\log_2 x + \log_3 y$ combines into nothing.
Multiply becomes add
One step down the operation ladder: $\times$ becomes $+$, $\div$ becomes $-$.
Derive, do not recall
Three lines rebuilds any law you have forgotten under exam pressure.
2
What You'll Master
objectives
Know
The product law $\log_a(xy) = \log_a x + \log_a y$
The quotient law $\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$
The power law $\log_a(x^n) = n \log_a x$
Understand
That each law is the matching index law with logarithm and index exchanged
Why all three require the same base, and why no law exists for $\log_a(x + y)$
Can Do
Derive each law from the index laws in three steps
State which index law a given logarithm law came from
Reject a false law by testing it on numbers
3
Words You Need
vocabulary
Product law$\log_a(xy) = \log_a x + \log_a y$, from $a^m \times a^n = a^{m+n}$.
Quotient law$\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$, from $a^m \div a^n = a^{m-n}$.
Power law$\log_a(x^n) = n\log_a x$, from $(a^m)^n = a^{mn}$.
DeduceTo obtain a result by reasoning from something already established, rather than by asserting it.
CounterexampleA single case that shows a claimed rule is false.
Common baseThe shared base every logarithm law requires before it can be applied.
4
Deriving the Product Law
+5 XP to read
The whole derivation is three moves. Watch where the index law does the work.
Name the logarithms. Let $m = \log_a x$ and $n = \log_a y$.
Translate to index form. By the definition, $a^m = x$ and $a^n = y$.
Apply the index law and translate back.
$$xy = a^m \times a^n = a^{m+n}$$
So $xy$ is $a$ raised to the power $m + n$, which by the definition means $\log_a(xy) = m + n$. Substituting back:
$$\log_a(xy) = \log_a x + \log_a y$$
Notice
The only mathematical step was $a^m \times a^n = a^{m+n}$. Everything else was translation.
5
Deriving the Quotient Law
+5 XP to read
Identical shape, one different index law.
Let $m = \log_a x$ and $n = \log_a y$, so $a^m = x$ and $a^n = y$. Then:
$$\frac{x}{y} = \frac{a^m}{a^n} = a^{m-n}$$
Therefore $\log_a\left(\dfrac{x}{y}\right) = m - n$, that is:
$$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$$
A useful special case falls straight out. Put $x = 1$, and since $\log_a 1 = 0$:
$$\log_a\left(\frac{1}{y}\right) = -\log_a y$$
6
Deriving the Power Law
+5 XP to read
This one needs only a single logarithm.
Let $m = \log_a x$, so $a^m = x$. Raise both sides to the power $n$:
$$x^n = (a^m)^n = a^{mn}$$
So $x^n$ is $a$ raised to the power $mn$, giving $\log_a(x^n) = mn$, that is:
$$\log_a(x^n) = n \log_a x$$
The index $n$ does not have to be a whole number. With $n = \dfrac{1}{2}$ this handles roots:
Set the index laws and the logarithm laws side by side and the pattern is unmistakable.
Index law
What the index does
Logarithm law
$a^m \times a^n = a^{m+n}$
added
$\log_a(xy) = \log_a x + \log_a y$
$a^m \div a^n = a^{m-n}$
subtracted
$\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$
$(a^m)^n = a^{mn}$
multiplied
$\log_a(x^n) = n\log_a x$
Each row moves the operation one rung down the ladder: multiplication becomes addition, division becomes subtraction, powers become multiplication. That is precisely what made logarithms worth inventing, three centuries before calculators.
8
Common Pitfalls
+5 XP to read
Writing $\log_a(x + y) = \log_a x + \log_a y$. There is no law for the logarithm of a sum.
Fix: test it. $\log_2(4 + 4) = \log_2 8 = 3$, but $\log_2 4 + \log_2 4 = 2 + 2 = 4$. One counterexample settles it.
Writing $\dfrac{\log_a x}{\log_a y} = \log_a x - \log_a y$. The quotient law is about the logarithm OF a quotient, not a quotient OF logarithms.
Fix: read the brackets. $\log_a\left(\dfrac{x}{y}\right)$ has the division inside; $\dfrac{\log_a x}{\log_a y}$ has it outside, and no law applies.
Applying a law across two different bases, such as $\log_2 8 + \log_3 9$.
Fix: every derivation began with a single $a$. With different bases there is no shared $a$ and nothing to combine, so evaluate each separately: $3 + 2 = 5$.
Watch Me Solve It · 3 examples
Watch Me Solve It · Deriving the product law from scratch
+15 XP per step
Q1
PROBLEM
Prove that $\log_a(xy) = \log_a x + \log_a y$, stating the index law you use.
1
Name the two logarithms
$\text{Let } m = \log_a x \text{ and } n = \log_a y$
Giving them names lets you work in index form.
2
Translate to index form
$a^m = x, \qquad a^n = y$
By the definition of a logarithm.
3
Multiply, and use the index law
$xy = a^m \times a^n = a^{m+n}$
The index law $a^m \times a^n = a^{m+n}$ is the only mathematical step.
4
Translate back
$\log_a(xy) = m + n = \log_a x + \log_a y$
Reading $a^{m+n} = xy$ as a logarithm returns the result.
Nice work, XP earned
Answer$\log_a(xy) = \log_a x + \log_a y$, from $a^m \times a^n = a^{m+n}$
Watch Me Solve It · Naming the parent index law
+15 XP per step
Q2
PROBLEM
The law $\log_a(x^5) = 5\log_a x$ follows from which index law? Show the connection in two lines.
1
Set the logarithm to a name
$\text{Let } m = \log_a x, \text{ so } a^m = x$
2
Raise both sides to the power 5
$x^5 = (a^m)^5 = a^{5m}$
This uses $(a^m)^n = a^{mn}$, the power-of-a-power law.
3
Read it as a logarithm
$\log_a(x^5) = 5m = 5\log_a x$
Nice work, XP earned
AnswerFrom $(a^m)^n = a^{mn}$
Watch Me Solve It · Disproving a false law
+15 XP per step
Q3
PROBLEM
A student claims $\log_a(x - y) = \log_a x - \log_a y$. Show that this is false.
1
Choose friendly numbers
$a = 2, \quad x = 8, \quad y = 4$
Pick values whose logarithms you know exactly, so no rounding can be blamed.
2
Evaluate the left-hand side
$\log_2(8 - 4) = \log_2 4 = 2$
3
Evaluate the right-hand side
$\log_2 8 - \log_2 4 = 3 - 2 = 1$
4
Conclude
$2 \neq 1$
A single counterexample is enough to disprove a claimed identity. The real quotient law concerns division, not subtraction.
$\log_5 5 = 1$, but $\log_5 2 + \log_5 3 = \log_5 6$, and $6 \neq 5$.No, there is no law for a sum
4 Simplify $\log_a\left(\dfrac{1}{x}\right)$.
Put $\dfrac{1}{x} = x^{-1}$ and use the power law.$-\log_a x$
Complete in your workbook.
Multiple Choice · 5 questions
MC1
The product law
+10 XP
$\log_a(xy)$ is equal to:
Correct, multiplying the numbers adds their indices, and a logarithm is an index.
Option A multiplies the logarithms, which corresponds to no index law at all. Option C is the quotient law.
Explanation: With $a^m = x$ and $a^n = y$, $xy = a^{m+n}$, so $\log_a(xy) = m + n = \log_a x + \log_a y$.
MC2
Naming the parent law
+10 XP
The power law $\log_a(x^n) = n\log_a x$ is deduced from:
Correct, raising a power to a power multiplies the indices, which is where the factor $n$ comes from.
Options A and B give the product and quotient laws. Option D gives $\log_a 1 = 0$.
Explanation: Writing $x = a^m$ and raising both sides to the power $n$ gives $x^n = a^{mn}$, so $\log_a(x^n) = mn = n\log_a x$.
MC3
A law that does not exist
+10 XP
Which of these statements is FALSE for all valid $a$, $x$ and $y$?
Correct, there is no law for the logarithm of a sum, and $\log_2(4+4) = 3$ while $\log_2 4 + \log_2 4 = 4$.
A, C and D are the product law, the power law with $n = 2$, and the power law with $n = -1$.
Explanation: Every logarithm law comes from an index law, and no index law simplifies $a^m + a^n$. That is exactly why no sum law exists.
MC4
Different bases
+10 XP
$\log_2 8 + \log_3 9$ equals:
Correct, the bases differ so no law applies, but each term evaluates on its own: $3 + 2 = 5$.
Options A and B apply the product law across two different bases, which none of the derivations permit. Option D confuses "no law applies" with "no value exists".
Explanation: $\log_2 8 = 3$ because $2^3 = 8$, and $\log_3 9 = 2$ because $3^2 = 9$. The sum is $5$. The laws only fail to COMBINE them; they are still perfectly evaluable.
MC5
Roots
+10 XP
$\log_a \sqrt{x}$ is equal to:
Correct, a square root is the index $\dfrac{1}{2}$, and the power law brings that index out in front.
Option C takes the root of the logarithm rather than the logarithm of the root, and option D invents a division.
Prove that $\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$. (a) Set out the proof in full, naming the index law you use. (b) Use your result to show that $\log_a\left(\dfrac{1}{y}\right) = -\log_a y$.
Write your working in your book.
(a) Let $m = \log_a x$ and $n = \log_a y$, so by the definition $a^m = x$ and $a^n = y$. Then $\dfrac{x}{y} = \dfrac{a^m}{a^n} = a^{m-n}$, using the index law $a^m \div a^n = a^{m-n}$. Reading that as a logarithm gives $\log_a\left(\dfrac{x}{y}\right) = m - n = \log_a x - \log_a y$.
(b) Put $x = 1$. Then $\log_a\left(\dfrac{1}{y}\right) = \log_a 1 - \log_a y$, and since $a^0 = 1$ gives $\log_a 1 = 0$, this is $-\log_a y$.
Marking guidance: 3 marks for (a): one for translating to index form, one for correctly naming and applying the division index law, one for translating back. 2 marks for (b): one for substituting $x = 1$ and one for using $\log_a 1 = 0$.
Q7
Test a claimed law
+15 XP
Q7
SHORT ANSWER
A student writes down three rules. For each, state whether it is true, and if it is false give a counterexample with actual numbers. (a) $\log_a(x^3) = 3\log_a x$ (b) $\dfrac{\log_a x}{\log_a y} = \log_a x - \log_a y$ (c) $\log_a(xy) = \log_a x + \log_a y$
Write your working in your book.
(a) True. It is the power law with $n = 3$, from $(a^m)^3 = a^{3m}$.
(b) False. Take $a = 2$, $x = 8$, $y = 4$. The left side is $\dfrac{3}{2} = 1.5$; the right side is $3 - 2 = 1$. The quotient law concerns the logarithm of a quotient, not a quotient of logarithms.
(c) True. It is the product law, from $a^m \times a^n = a^{m+n}$.
Marking guidance: 1 mark each for correctly classifying (a) and (c). 2 marks for (b): one for identifying it as false and one for a valid numerical counterexample.
Q8
Why there is no sum law
+15 XP
Q8
SHORT ANSWER
Explain, by reference to the index laws, why there is no law that simplifies $\log_a(x + y)$. Support your explanation with a numerical counterexample to the claim $\log_a(x + y) = \log_a x + \log_a y$.
Write your working in your book.
Every logarithm law is obtained by writing $x = a^m$ and $y = a^n$ and then applying an index law to the result. The three index laws describe what happens when powers are multiplied, divided, or raised to a power; none of them simplifies $a^m + a^n$, because adding two powers of the same base does not in general produce a power of that base. With no index law to translate, there is no logarithm law to obtain.
Counterexample: with $a = 2$, $x = 4$ and $y = 4$, the left side is $\log_2 8 = 3$ while the right side is $\log_2 4 + \log_2 4 = 4$. Since $3 \neq 4$ the claimed rule is false.
Marking guidance: 2 marks for the explanation, one for connecting each logarithm law to a parent index law and one for observing that no index law handles a sum of powers. 1 mark for a correct counterexample with both sides evaluated.
S
Stretch Challenge · Why logarithms were invented
+25 XP
S
CHALLENGE
Before calculators, astronomers multiplied large numbers using printed tables of common logarithms. (a) Explain, using the product law, how a table of logarithms turns a multiplication into an addition. (b) Using $\log_{10} 2 \approx 0.3010$ and $\log_{10} 3 \approx 0.4771$, find $\log_{10} 6$ without multiplying. (c) Extend the method: find $\log_{10} 72$ from the same two values alone, showing every law you use.
(a) To find $xy$, look up $\log_{10} x$ and $\log_{10} y$, add them to get $\log_{10}(xy)$ by the product law, then read the table backwards to recover $xy$. The multiplication is replaced by an addition and two lookups, which is far faster and far less error-prone by hand.