Logarithmic and Exponential Functions Revision Notes — 9709 Paper 3

Logarithms and the index laws

Syllabus requirement · §3.2

“

understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base).

”

Why we need logarithms

Solve $2^x = 8$ . You can do it by inspection: $2^3 = 8$ , so $x = 3$ . Now solve $2^x = 5$ . No whole number works: $2^2 = 4$ is too small and $2^3 = 8$ is too big, so $x$ is somewhere between $2$ and $3$ . We need a name for "the power of $2$ that gives $5$ ", and a way to calculate it. That name is a logarithm.

What a logarithm is

A logarithm answers one question: what power do I need?

\log_a c = b \quad\text{means exactly the same as}\quad a^b = c

$a$ is the base. It is a positive number other than $1$ .
$c$ is the number you want to reach.
$b$ , the logarithm, is the power you must raise $a$ to in order to reach $c$ .

Read $\log_2 8 = 3$ aloud as "log base $2$ of $8$ is $3$ ". It says the same thing as $2^3 = 8$ . So the answer to $2^x = 5$ is simply $x = \log_2 5$ , and later in this note you will see how to evaluate it on a calculator.

The two forms are the same fact written two ways. Being able to switch between them on sight is the foundation of the whole topic.

Index form	Log form	Read it as
$2^3 = 8$	$\log_2 8 = 3$	the power of $2$ that gives $8$ is $3$
$10^{-2} = 0.01$	$\log_{10} 0.01 = -2$	a log can be negative
$9^{1/2} = 3$	$\log_9 3 = \tfrac12$	a log can be a fraction
$5^0 = 1$	$\log_5 1 = 0$	the log of $1$ is always $0$
$7^1 = 7$	$\log_7 7 = 1$	the log of the base is always $1$

The same five facts in both forms. Cover one column and rebuild it from the other.

You cannot take the log of zero or a negative number

With a positive base, $a^b$ is positive for every power $b$ : $2^{10} = 1024$ , $2^{-10} = \tfrac{1}{1024}$ , and no power of $2$ is ever $0$ or negative. So there is no power that gives $-4$ or $0$ , and $\log_a(-4)$ and $\log_a 0$ do not exist.

This matters later: when you solve an equation containing $\log_a(\ldots)$ , every bracket inside a log must be positive. Any root that makes one zero or negative has to be thrown away.

Evaluating logs by switching to index form

Without a calculator, find (a) $\log_3 81$ (b) $\log_2\!\left(\tfrac18\right)$ (c) $\log_{25} 5$ .

Show full working

1
(a) Call the answer $b$ . By the definition, $\log_3 81 = b$ means $3^b = 81$
Every one of these starts the same way: name the unknown power, then rewrite the log as an index statement.
2
Write $81$ as a power of $3$ : $81 = 3 \times 3 \times 3 \times 3 = 3^4$ . So $3^b = 3^4$ , and $b = 4$ .
3
(b) $\log_2\!\left(\tfrac18\right) = b$ means $2^b = \tfrac18$
4
$8 = 2^3$ , and a reciprocal is a negative power, so $\tfrac18 = 2^{-3}$ . Hence $b = -3$ .
The index law $a^{-n} = \dfrac{1}{a^n}$ is what turns "one over" into a negative log.
5
(c) $\log_{25} 5 = b$ means $25^b = 5$
6
$5$ is the square root of $25$ , and a square root is the power $\tfrac12$ : $25^{1/2} = 5$ . Hence $b = \tfrac12$ .

Answer

(a) $4$ (b) $-3$ (c) $\tfrac12$

If you cannot see the value of a log, write it as an index equation. The question becomes "what power of the base gives this number?", which you can usually answer.

The three laws

\log_a(xy) = \log_a x + \log_a y

Product → sum

\log_a\!\left(\tfrac{x}{y}\right) = \log_a x - \log_a y

Quotient → difference

\log_a\!\left(x^{\,n}\right) = n\log_a x

Power → multiplier (the one that does the work)

Where the laws come from

Each law is an index law written in log form. Let $\log_a x = m$ and $\log_a y = n$ . By the definition, that means

x = a^m \qquad\text{and}\qquad y = a^n

Product law. Multiply $x$ and $y$ , and use $a^m \times a^n = a^{m+n}$ :

xy = a^m \times a^n = a^{m+n}

Written in log form, this says the power of $a$ that gives $xy$ is $m + n$ :

\log_a(xy) = m + n = \log_a x + \log_a y

Quotient law. Divide instead, and use $a^m \div a^n = a^{m-n}$ :

\frac{x}{y} = a^{m-n} \quad\Longrightarrow\quad \log_a\!\left(\frac{x}{y}\right) = m - n = \log_a x - \log_a y

Power law. Raise $x$ to the power $k$ , and use $(a^m)^k = a^{mk}$ :

x^k = \left(a^m\right)^k = a^{km} \quad\Longrightarrow\quad \log_a\!\left(x^k\right) = km = k\log_a x

So you never need to memorise the log laws separately. If one ever looks doubtful, translate it back into indices and check.

The last two rows of the table above are also index laws: $a^0 = 1$ gives $\log_a 1 = 0$ , and $a^1 = a$ gives $\log_a a = 1$ .

Base e, base 10, base anything

Your calculator has two log buttons. $\log$ is base $10$ , written $\log_{10}$ . $\ln$ is base $\mathrm{e}$ , where $\mathrm{e} \approx 2.718$ is a special number explained in the next section. So $\ln x$ means $\log_{\mathrm{e}} x$ , and from here on most examples use $\ln$ .

The laws hold in every base. When you take logs of both sides of an equation you may use $\ln$ or $\log_{10}$ , as long as you use the same base throughout. Mark schemes say exactly this: "Allow logs to any base but must be consistent throughout."

Changing a log from one base to another is not on the syllabus. When a question is set in $\log_3$ or $\log_4$ , work in that base all the way through.

Using the laws in both directions

Combining several logs into one is how you solve log equations. Splitting one log into several (or recombining at the end) is how you reach the exact answers questions ask for.

Adding logs multiplies the insides, subtracting divides them, and a number in front becomes a power:

\ln 4 + \ln 5 = \ln 20 \qquad \ln 20 - \ln 4 = \ln 5 \qquad 3\ln 2 = \ln 2^3 = \ln 8

The number in front always moves first. In $2\ln 3 + \ln 5$ the product law cannot act yet, because it only combines logs that stand alone. First $2\ln 3 = \ln 3^2 = \ln 9$ , and only then $\ln 9 + \ln 5 = \ln 45$ . Writing $2\ln 3 + \ln 5 = \ln 15$ is a very common slip.

Combining logs into a single log

Write $2\ln 6 - \ln 4 - \ln 3$ as a single logarithm in its simplest form.

Show full working

1
Move the number in front inside first, using the power law: $2\ln 6 = \ln 6^2 = \ln 36$
The product and quotient laws only combine bare logs. Any multiplier has to become a power before anything else happens.
2
The expression is now $\ln 36 - \ln 4 - \ln 3$
3
Quotient law on the first two terms: $\ln 36 - \ln 4 = \ln\!\left(\frac{36}{4}\right) = \ln 9$
4
Quotient law again: $\ln 9 - \ln 3 = \ln\!\left(\frac{9}{3}\right) = \ln 3$
Each minus sign divides by the next inside. Both 4 and 3 end up in the denominator: 36 ÷ (4 × 3) = 3.

Answer

$\ln 3$

Powers first, then combine. Every term with a minus sign in front ends up in the denominator.

The three that are not laws

These look plausible and are all false:

\log(x+y) \;\neq\; \log x + \log y \qquad \frac{\log x}{\log y} \;\neq\; \log\!\left(\frac{x}{y}\right) \qquad (\log x)^2 \;\neq\; 2\log x

The laws convert products into sums, never sums into sums. A quick numerical check settles any doubt: $\log_{10}(10+90) = 2$ , but $\log_{10}10 + \log_{10}90 \approx 2.95$ .

Note also the third one carefully: $\log(x^2) = 2\log x$ , but $(\log x)^2$ is the log squared and nothing simplifies.

Logs of letters

The laws work just the same when the insides are letters. $\ln(p^2q) = 2\ln p + \ln q$ , for example. When a question gives you facts about $\ln p$ and $\ln q$ , split every log down to $\ln p$ and $\ln q$ and treat those two as the unknowns of a pair of simultaneous equations.

Expressing one log in terms of two others

9709/32 F/M 2024 Q44 marks

The positive numbers $p$ and $q$ are such that

\ln\left(\frac{p}{q}\right) = a \quad \text{and} \quad \ln(q^2p) = b.

Express $\ln(p^7q)$ in terms of $a$ and $b$ .

Show full working

1
Quotient law on the first fact: $\ln p - \ln q = a$
Each of these two split-up equations is a B1.
2
Product law, then power law, on the second fact: $\ln(q^2p) = \ln q^2 + \ln p = 2\ln q + \ln p = b$
3
Treat $\ln p$ and $\ln q$ as two unknowns. Subtract the first equation from the second: $(\ln p + 2\ln q) - (\ln p - \ln q) = b - a$
4
The $\ln p$ terms cancel, and $2\ln q - (-\ln q) = 3\ln q$ : $3\ln q = b - a \;\Longrightarrow\; \ln q = \frac{b - a}{3}$
5
From the first equation, $\ln p = a + \ln q$ : $\ln p = a + \frac{b-a}{3} = \frac{3a + b - a}{3} = \frac{2a + b}{3}$
6
Split the target the same way: $\ln(p^7q) = \ln p^7 + \ln q = 7\ln p + \ln q$
7
Substitute: $7\left(\frac{2a+b}{3}\right) + \frac{b-a}{3} = \frac{14a + 7b + b - a}{3}$
8
Collect like terms: $\ln(p^7q) = \frac{13a + 8b}{3}$

Answer

$\ln(p^7q) = \dfrac{13a + 8b}{3}$

Split every log into ln p and ln q first. After that it is ordinary simultaneous equations.

Your turn

Every later section assumes you can do these quickly, in either direction, without a calculator.

1
Find the value of $\log_4 8$ without a calculator.
Stuck? Show hint
Write $\log_4 8 = b$ as $4^b = 8$ , then write both $4$ and $8$ as powers of $2$ .
Show solution
1. 1
  Let $\log_4 8 = b$ . By the definition: $4^b = 8$
2. 2
  $8$ is not a whole-number power of $4$ , but both are powers of $2$ : $4 = 2^2$ and $8 = 2^3$ . So $\left(2^2\right)^b = 2^3$
3. 3
  Multiply the powers on the left: $2^{2b} = 2^3$
  The index law $(a^m)^n = a^{mn}$ .
4. 4
  Both sides are powers of $2$ , so the powers must be equal: $2b = 3 \;\Longrightarrow\; b = \tfrac32$
Answer
$\log_4 8 = \tfrac32$
2
Write $3\ln 2 + \ln 5 - \ln 4$ as a single logarithm.
Stuck? Show hint
Move every number in front inside first, using the power law, before you combine anything.
Show solution
1. 1
  Power law on the first term: $3\ln 2 = \ln 2^3 = \ln 8$
  Multipliers go inside first. Only bare logs can be combined.
2. 2
  Product law on the first two terms: $\ln 8 + \ln 5 = \ln(8 \times 5) = \ln 40$
3. 3
  Quotient law with the last term: $\ln 40 - \ln 4 = \ln\!\left(\frac{40}{4}\right) = \ln 10$
Answer
$\ln 10$
3
Given that $\log_a 2 = p$ and $\log_a 5 = q$ , express $\log_a 20$ in terms of $p$ and $q$ .
Stuck? Show hint
Split 20 into factors you already have logs for — 20 = 4 × 5 = 2² × 5.
Show solution
1. 1
  Write $20 = 2^2 \times 5$ , so $\log_a 20 = \log_a\!\left(2^2 \times 5\right)$
2. 2
  Product law: $= \log_a 2^2 + \log_a 5$
3. 3
  Power law on the first term: $= 2\log_a 2 + \log_a 5$
4. 4
  Substitute $\log_a 2 = p$ and $\log_a 5 = q$ : $= 2p + q$
  Splitting a number into factors you already know the logs of is the same move, run backwards, that produces exact answers such as ln 24 = 3 ln 2 + ln 3 later on.
Answer
$2p + q$
4
The positive numbers $x$ and $y$ satisfy $\ln(xy^2) = 7$ and $\ln\!\left(\dfrac{x}{y}\right) = 1$ . Find the values of $\ln x$ and $\ln y$ , and hence find the value of $\ln(x^3y)$ .
Stuck? Show hint
Split both facts into $\ln x$ and $\ln y$ , then solve the pair of simultaneous equations.
Show solution
1. 1
  Product and power laws on the first fact: $\ln x + 2\ln y = 7$
2. 2
  Quotient law on the second: $\ln x - \ln y = 1$
3. 3
  Subtract the second from the first. The $\ln x$ terms cancel: $3\ln y = 6 \;\Longrightarrow\; \ln y = 2$
4. 4
  Substitute into the second: $\ln x = 1 + \ln y = 3$
5. 5
  Split the target: $\ln(x^3y) = 3\ln x + \ln y = 3(3) + 2 = 11$
Answer
$\ln x = 3$ , $\ln y = 2$ , $\ln(x^3y) = 11$

Checking your access…

Can you do all of these?

Convert between aᵇ = c and log_a c = b, and evaluate simple logs without a calculator
State the three log laws, move a number in front inside first, and know that log(x + y) does not split
Use ln and e to undo each other; simplify $\mathrm{e}^{2\ln 5}$ by moving the 2 inside first
Sketch $y = \mathrm{e}^{kx}$ for k positive and k negative, and a shifted or reflected version, with asymptote and intercepts
Explain why eˣ and ln x are reflections in y = x, and give the domain of ln x
Use $Q = A\mathrm{e}^{kt}$ : $A$ is the starting value; solve $\mathrm{e}^{kt} = 2$ or $\tfrac12$ for the doubling or halving time
Split logs of letters into ln p and ln q and solve for them as simultaneous equations
Use two sketched curves to show an equation has exactly one root
Solve an index equation with different bases, keeping brackets on the powers
Simplify a same-base equation with index laws before taking any log
Spot a hidden quadratic in aˣ or eˣ, and reject a non-positive value of aˣ with a reason
Give an exact answer as ln a / ln b by recombining logs at the end
Solve an index inequality, reversing the sign when dividing by the log of a number below 1
Split a modulus equation in aˣ into two cases and keep every valid answer
Write down the domain, combine logs into a single log, remove it, and check every root
Absorb a number or a term as a log: 2 = log₄ 16, 2x = ln e²ˣ
Make a named letter the subject of a log relationship
Reduce a model to linear form, say what is plotted against what, and compare explicitly with Y = mX + c
Convert a gradient of ln a back into a, and use ln x = 0 (not x = 0) for the vertical-axis intercept

Now do the questions

59 real Paper 3 parts from 2021–2025, sorted by difficulty, with mark schemes

Algebra

Trigonometry

Logarithms and the index laws

Why we need logarithms

What a logarithm is

You cannot take the log of zero or a negative number

Where the laws come from

Using the laws in both directions

Logs of letters

The rest of this note

Can you do all of these?