Maths · 3 Functions

3.7 The logarithmic function

Before you start

Inverse functions and what it means for one function to undo another (3.3).
The shape of an exponential graph and that it passes through $(0,1)$ (3.1).
Using the log button on a graphic display calculator.

Syllabus objectives covered (1)

Understand and use: • the logarithmic function as the inverse of the exponential function • y = aˣ as equivalent to x = logₐ y • the solution to aˣ = b as x = (log b)/(log a).

Part 1 of 3

A logarithm answers a question about a power

A logarithm asks: to what power must I raise this base to get that number? That sentence is the whole idea, and everything else follows from it.
So $\log_{2} 8 = 3$ , because you must raise 2 to the power 3 to get 8. Read it as 'log, base 2, of 8'.
The two statements below say exactly the same thing, written two ways:
$y = a^{x} ⟺ x = \log_{a} y$
Being able to move between those forms is most of what this subtopic asks of you. Practise converting in both directions until it is automatic.
This is why the log is the inverse of the exponential function. If $f (x) = a^{x}$ then $f^{- 1} (x) = \log_{a} x$ — one raises to a power, the other retrieves the power.
The syllabus notes say all logs are base 10 unless otherwise stated, so a plain $\log$ means $\log_{10}$ . That is the `log` button on your calculator.
Two values are worth knowing on sight: $\log 1 = 0$ , because $10^{0} = 1$ ; and $\log 10 = 1$ , because $10^{1} = 10$ .

Part 2 of 3

Solving an exponential equation

The reason logarithms are on this syllabus is that they let you get at an unknown exponent, which ordinary algebra cannot reach.
The syllabus gives you the result directly:
$a^{x} = b ⟹ x = \frac{\log b}{\log a}$
Note carefully: it is $\log b$ divided by $\log a$ , not $\log (b / a)$ and not $\log b - \log a$ . Those are different and both are wrong here.
For $3^{x} = 20$ : $x = \frac{\log 20}{\log 3} = \frac{1.301}{0.4771} = 2.73$ to 3 significant figures.
Check by substituting back: $3^{2.73} \approx 20.0$ . Doing this once per question catches most calculator-entry errors.
If the equation is not already in the form $a^{x} = b$ , rearrange it first. For $5 \times 2^{x} = 80$ , divide by 5 to get $2^{x} = 16$ , and only then apply the rule.
When the answer comes out exact, say so: $2^{x} = 16$ gives $x = 4$ exactly, and there is no need to write $4.00$ .

Part 3 of 3

Growth, decay, and interpreting the answer

The syllabus notes point at where these equations come from: compound interest, and exponential growth and decay problems.
Compound interest at $r %$ per year on an initial amount $P$ gives, after $n$ years:
$A = P {(1 + \frac{r}{100})}^{n}$
Growth means the multiplier is greater than 1; decay means it is between 0 and 1. A 15% annual fall gives a multiplier of $0.85$ , not $- 0.15$ .
To find when something reaches a certain value, set up the equation, isolate the exponential part, and use the log rule.
Then interpret the answer in context, which is where marks are most often lost. A calculator value of $7.4$ years usually means 8 whole years, because at 7 years the target has not yet been reached.
So read the question's wording carefully: 'after how many complete years' means round up, not to the nearest.
You can also solve these on the graphic display calculator by graphing both sides and finding the intersection, as in 3.2. Either method is acceptable — but show what you did.

Worked examples

Example 1. Solve $4^{x} = 30$ , giving your answer correct to 3 significant figures.

Show the working

The equation is already in the form $a^{x} = b$ , with $a = 4$ and $b = 30$ .
Apply the rule from the syllabus.
$x = \frac{\log 30}{\log 4}$
Evaluate on the calculator without rounding the two logs separately.
$x = \frac{1.4771}{0.6021} = 2.4534 \dots$
Round only at the end, to 3 significant figures: $x = 2.45$ .
Check: $4^{2.45} \approx 29.9$ , close to 30. And a sanity check on size — $4^{2} = 16$ and $4^{3} = 64$ , so the answer must lie between 2 and 3. It does.

Answer: $x = 2.45$ (3 s.f.).

Example 2. $$ 2000$ is invested at $6 %$ compound interest per year. Find the number of complete years before the investment first exceeds $$ 3000$ .

Show the working

Set up the compound interest equation with a multiplier of $1.06$ per year.
$2000 \times {1.06}^{n} = 3000$
Isolate the exponential part by dividing both sides by 2000.
${1.06}^{n} = 1.5$
Now it is in the form $a^{x} = b$ , so apply the log rule.
$n = \frac{\log 1.5}{\log 1.06} = \frac{0.17609}{0.025306} = 6.958 \dots$
Interpret in context. After 6 years the investment has not yet reached $$ 3000$ ; it passes it partway through year 7.
So the answer is 7 complete years. Rounding $6.96$ to the nearest whole number happens to give 7 here, but the reasoning — round up — is what matters and is what earns the mark.

Answer: $7$ complete years.

Example 3. Write $\log_{5} 125 = 3$ in exponential form, and write $2^{6} = 64$ in logarithmic form.

Show the working

Use the equivalence $y = a^{x} \Leftrightarrow x = \log_{a} y$ , and identify the parts each time.
In $\log_{5} 125 = 3$ , the base is 5, the answer (the power) is 3, and 125 is the result.
So in exponential form: $5^{3} = 125$ . Check: $5 \times 5 \times 5 = 125$ . Correct.
In $2^{6} = 64$ , the base is 2, the power is 6, and 64 is the result.
So in logarithmic form the power becomes the answer: $\log_{2} 64 = 6$ .

Answer: $5^{3} = 125$ , and $\log_{2} 64 = 6$ .

Where people go wrong

Mistake: Writing $\frac{\log b}{\log a}$ as $\log (\frac{b}{a})$ or as $\log b - \log a$ .

Instead: It is a division of two separate logarithms, not a log of a quotient.

$\log 20 / \log 3 = 2.73$ , whereas $\log (20 / 3) = 0.824$ . Completely different answers.

Mistake: Using a multiplier of $- 0.15$ for a 15% decrease.

Instead: The multiplier is $0.85$ .

Each year the quantity becomes 85% of what it was. A negative multiplier would make the value alternate in sign, which is meaningless here.

Mistake: Rounding an answer like $6.958$ to $7$ by habit, without checking the context.

Instead: Ask whether the question wants the value rounded, or the first whole year past the target.

'Complete years' means round up. A value of $7.1$ would round to 7 but the correct answer would be 8.

Mistake: Rounding $\log b$ and $\log a$ before dividing them.

Instead: Do the whole calculation in one go on the calculator.

The syllabus accuracy rules require it, and rounding two intermediate values can shift the final digit.

Mistake: Applying the log rule before isolating the exponential term.

Instead: Rearrange first so the equation reads $a^{x} = b$ with nothing else attached.

In $5 \times 2^{x} = 80$ , taking logs immediately makes the algebra far harder. Divide by 5 first.

Check yourself

Write $\log_{3} 81 = 4$ in exponential form.
Answer
$3^{4} = 81$ .
What is the value of $\log 1$ , and why?
Answer
$0$ , because $10^{0} = 1$ .
Solve $2^{x} = 50$ , correct to 3 significant figures.
Answer
$x = \frac{\log 50}{\log 2} = 5.64$ .
Solve $3 \times 4^{x} = 48$ .
Answer
Divide by 3 to get $4^{x} = 16$ , so $x = 2$ exactly.
What is the multiplier for a quantity that decreases by 8% each year?
Answer
$0.92$ .
Explain why the logarithmic function is the inverse of the exponential function.
Answer
An exponential raises a base to a power; a logarithm returns the power that produced a given result. Each undoes the other.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A population of bacteria is modelled by $N = 500 \times {1.12}^{t}$ , where $t$ is the time in hours.

(a) Work out the population after $5$ hours, correct to the nearest whole number. 2 marks

Mark scheme

M1 500 × 1.12⁵ seen oe
A1 881 cao

(b) Find the time taken for the population to reach $2000$ , giving your answer correct to 3 significant figures. 3 marks

Mark scheme

M1 1.12^t = 4 seen oe
M1 t = log 4 / log 1.12 oe
A1 12.2 awrt

(c) Write down the number of complete hours after which the population first exceeds $2000$ . 1 mark No calculator

Mark scheme

B1 13; FT their (b) rounded up

Question 2. Two statements are given: $\log_{a} 64 = 2$ and $7^{x} = 300$ .

(a) Find the value of $a$ . 2 marks No calculator

Mark scheme

M1 a² = 64 seen
A1 a = 8

(b) Solve $7^{x} = 300$ , giving your answer correct to 3 significant figures. 2 marks

Mark scheme

M1 x = log 300 / log 7 oe
A1 2.93 awrt

Question 3. A car bought for $$ 24 000$ loses $18 %$ of its value each year.

(a) Write down the multiplier for one year's depreciation. 1 mark No calculator

Mark scheme

B1 0.82

(b) Find the number of complete years after which the car is first worth less than $$ 10 000$ . 4 marks

Mark scheme

M1 24000 × 0.82ⁿ = 10000 formed; FT their multiplier
M1 0.82ⁿ = 0.41666… oe
M1 n = log 0.41666… / log 0.82 oe
A1 5 (complete years); dep on a value between 4 and 5 seen

Words to know

Logarithm: The power to which a base must be raised to give a number: $\log_{a} b = x$ means $a^{x} = b$ .
Base: The number being raised to a power. Assume base 10 unless stated otherwise.
Exponential function: A function of the form $f (x) = a^{x}$ , whose inverse is $\log_{a} x$ .
Multiplier: The number a quantity is multiplied by each period: $1.06$ for a 6% rise, $0.82$ for an 18% fall.
Compound interest: Interest calculated on the running total, giving $A = P {(1 + r / 100)}^{n}$ .

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