Cambridge IGCSE Mathematics 0607 · Extended · Syllabus 2025-2027

Logarithms Worked ExamplesIGCSE Maths 0607

Exam-style worked examples on logarithms — evaluating them from the definition, the three log laws, and solving exponential equations, including graph-based questions on $y = \log_2 x$ as a reflection of $y = 2^x$ and exponential models read from graphs. Note that logarithms are assessed in International Mathematics 0607 rather than in 0580, so check which syllabus you are entered for.

This page has 9 exam-style IGCSE Maths 0607 logarithms worked examples — 1 Core, 6 Extended and 2 Challenge, 3 with diagrams. Each one shows the full method step by step, the final answer in the form a marker expects, and an examiner tip on where marks are commonly lost.

Example 1CoreEvaluating logarithms[3 marks]

Logarithms from the definition

Without a calculator, find the value of:

(a) $\log_{10} 1000$ (b) $\log_2 32$ (c) $\log_5 1$

Example 2ExtendedLog laws[4 marks]

Combining logarithms

Write each as a single logarithm, then evaluate:

(a) $\log_{10} 4 + \log_{10} 25$ (b) $\log_{10} 200 - \log_{10} 2$

Example 3ExtendedLog laws[3 marks]

The power law

(a) Show that $3\log_{10} 2 = \log_{10} 8$ . (b) Hence simplify $2\log_{10} 3 + \log_{10} 2$ to a single logarithm.

Example 4ExtendedExponential equations[4 marks]

Solving an exponential equation

Solve $2^x = 20$ , giving your answer correct to $3$ significant figures.

Example 5ExtendedLog equations[4 marks]

Solving a logarithmic equation

Solve $\log_{10} x + \log_{10} 4 = 2$ .

Example 6ChallengeExponential equations[5 marks]

Exponential growth with logarithms

A population of $5000$ grows by $6\%$ each year, so after $n$ years it is $P = 5000 \times 1.06^{n}$ .

Use logarithms to find the smallest whole number of years for the population to exceed $8000$ .

Example 7ExtendedLogarithm graph as the inverse of an exponential[6 marks]

The graph of a logarithm as a reflection

The diagram shows the graph of $y = 2^x$ and its reflection in the line $y = x$ , which is the graph of $y = \log_2 x$ .

(a) The points $(1, 2)$ , $(2, 4)$ and $(3, 8)$ are marked on $y = 2^x$ . Write down the coordinates of the three matching points on $y = \log_2 x$ , and explain your answer.

(b) Use the graph of $y = \log_2 x$ to estimate $\log_2 6$ . Check your estimate with a calculator, using $\log_2 6 = \dfrac{\log 6}{\log 2}$ .

(c) Use the graph to solve $\log_2 x = -1$ .

(d) Explain, using the graph of $y = 2^x$ , why $\log_2 x$ has no value when $x \leqslant 0$ .

Example 8ExtendedCalculator: exponential model from a graph[8 marks]

Exponential model from a graph, solved with logarithms

The number of bacteria, $N$ , in a culture $t$ hours after the start of an experiment is modelled by $N = a \times b^{t}$ . The graph of $N$ against $t$ is shown.

(a) Use the two points marked on the graph to find the values of $a$ and $b$ .

(b) Use the graph to estimate the time when $N = 2000$ . Then use logarithms to find this time correct to $3$ significant figures.

(c) The experiment stops when $N$ reaches $5000$ . Use logarithms to find when this happens. Give your answer in hours and minutes, correct to the nearest minute.

1
(a) At $t = 0$ the power is $b^0 = 1$ , so the starting value gives $a$ directly.
$400 = a \times b^0 = a \quad \Rightarrow \quad a = 400$
2
Substitute the second point. $b$ must be positive for a growth model.
$900 = 400b^2 \quad \Rightarrow \quad b^2 = 2.25 \quad \Rightarrow \quad b = 1.5$
3
(b) Go across from $N = 2000$ to the curve, then down to the $t$ -axis. The reading is just under $4$ hours.
4
Set up the equation, divide by $400$ , then take logs of both sides.
$400 \times 1.5^{t} = 2000 \;\Rightarrow\; 1.5^{t} = 5 \;\Rightarrow\; t\log 1.5 = \log 5$
5
Divide. The answer agrees with the graph. ✓
$t = \frac{\log 5}{\log 1.5} = 3.969\ldots \approx 3.97 \text{ hours}$
6
(c) $N = 5000$ is beyond the graph, so logarithms are the only accurate method.
$400 \times 1.5^{t} = 5000 \;\Rightarrow\; 1.5^{t} = 12.5 \;\Rightarrow\; t = \frac{\log 12.5}{\log 1.5} = 6.2292\ldots$
7
Convert the decimal part of an hour to minutes.
$0.2292\ldots \times 60 = 13.75\ldots \approx 14 \text{ minutes}$

Answer

(a) $a = 400$ , $b = 1.5$ (b) about $4$ hours from the graph; $3.97$ hours (c) $6$ hours $14$ minutes

Examiner tip

In (c), $6.23$ hours is not $6$ hours $23$ minutes — multiply the decimal part by $60$ , and keep it unrounded until then. $\frac{\log 12.5}{\log 1.5}$ is a division of two logs, not $\log\left(\frac{12.5}{1.5}\right)$ . Always compare the logarithm answer with the graph reading; a mismatch means a slip.

Example 9ChallengeShow that: comparing two exponential models[8 marks]

When one population overtakes another

The populations of two towns are modelled by $P_A = 500 \times 1.08^{t}$ and $P_B = 900 \times 1.03^{t}$ , where $t$ is the time in years. The graphs of both models are shown.

(a) Use the graph to estimate the value of $t$ when the two populations are equal.

(b) Show that this value of $t$ satisfies $t = \dfrac{\log 1.8}{\log 1.08 - \log 1.03}$ .

(c) Find this value of $t$ correct to $3$ significant figures, and the populations at that time, correct to the nearest $10$ .

(d) Find the smallest whole number of years after which the population of town $A$ is greater than the population of town $B$ .

1
(a) The curves cross between $t = 12$ and $t = 13$ , a little under halfway, so $t \approx 12.4$ .
2
(b) Set the two models equal.
$500 \times 1.08^{t} = 900 \times 1.03^{t}$
3
Gather the powers on one side: divide both sides by $500$ and by $1.03^{t}$ .
$\frac{1.08^{t}}{1.03^{t}} = \frac{900}{500} = 1.8$
4
Take logs. The division law turns the left side into a difference.
$t\log 1.08 - t\log 1.03 = \log 1.8$
5
Factorise out $t$ and divide.
$t(\log 1.08 - \log 1.03) = \log 1.8 \;\Rightarrow\; t = \frac{\log 1.8}{\log 1.08 - \log 1.03}$
6
(c) Evaluate, keeping full accuracy. This agrees with the graph. ✓
$t = \frac{0.25527\ldots}{0.020586\ldots} = 12.39\ldots \approx 12.4$
7
Substitute the unrounded $t$ into either model — both give the same value.
$500 \times 1.08^{12.39\ldots} = 1298.4\ldots \approx 1300$
8
(d) Test the whole numbers either side of the crossing. After $12$ years $A$ is still smaller; after $13$ years it is larger.
$t = 12: \; 1259 < 1283, \qquad t = 13: \; 1360 > 1322$

Answer

(a) about $12.4$ (b) shown (c) $t = 12.4$ ; both populations about $1300$ (d) $13$ years

Examiner tip

The key step in (b) is getting both powers of $t$ together before taking logs. A common error is to write $\log(500 \times 1.08^{t})$ as $500t\log 1.08$ — the $500$ is multiplied, so it becomes $\log 500 + t\log 1.08$ . In (d), test whole numbers either side of the crossing rather than just rounding $12.4$ .

These worked examples are written by Tutopiya to cover the same syllabus content as the official Cambridge IGCSE 0607 specification (2025-2027). They are exam-style questions, not reproductions of past paper questions.

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← Back to Logarithms study notes

Logarithms Worked ExamplesIGCSE Maths 0607

Example 1CoreEvaluating logarithms[3 marks]

Logarithms from the definition

Without a calculator, find the value of:

(a) $\log_{10} 1000$ (b) $\log_2 32$ (c) $\log_5 1$

Example 2ExtendedLog laws[4 marks]

Combining logarithms

Write each as a single logarithm, then evaluate:

(a) $\log_{10} 4 + \log_{10} 25$ (b) $\log_{10} 200 - \log_{10} 2$

Example 3ExtendedLog laws[3 marks]

The power law

(a) Show that $3\log_{10} 2 = \log_{10} 8$ . (b) Hence simplify $2\log_{10} 3 + \log_{10} 2$ to a single logarithm.

Example 4ExtendedExponential equations[4 marks]

Solving an exponential equation

Solve $2^x = 20$ , giving your answer correct to $3$ significant figures.

Example 5ExtendedLog equations[4 marks]

Solving a logarithmic equation

Solve $\log_{10} x + \log_{10} 4 = 2$ .

Example 6ChallengeExponential equations[5 marks]

Exponential growth with logarithms

A population of $5000$ grows by $6\%$ each year, so after $n$ years it is $P = 5000 \times 1.06^{n}$ .

Use logarithms to find the smallest whole number of years for the population to exceed $8000$ .

Example 7ExtendedLogarithm graph as the inverse of an exponential[6 marks]

The graph of a logarithm as a reflection

The diagram shows the graph of $y = 2^x$ and its reflection in the line $y = x$ , which is the graph of $y = \log_2 x$ .

(a) The points $(1, 2)$ , $(2, 4)$ and $(3, 8)$ are marked on $y = 2^x$ . Write down the coordinates of the three matching points on $y = \log_2 x$ , and explain your answer.

(b) Use the graph of $y = \log_2 x$ to estimate $\log_2 6$ . Check your estimate with a calculator, using $\log_2 6 = \dfrac{\log 6}{\log 2}$ .

(c) Use the graph to solve $\log_2 x = -1$ .

(d) Explain, using the graph of $y = 2^x$ , why $\log_2 x$ has no value when $x \leqslant 0$ .

Example 8ExtendedCalculator: exponential model from a graph[8 marks]

Exponential model from a graph, solved with logarithms

The number of bacteria, $N$ , in a culture $t$ hours after the start of an experiment is modelled by $N = a \times b^{t}$ . The graph of $N$ against $t$ is shown.

(a) Use the two points marked on the graph to find the values of $a$ and $b$ .

(b) Use the graph to estimate the time when $N = 2000$ . Then use logarithms to find this time correct to $3$ significant figures.

(c) The experiment stops when $N$ reaches $5000$ . Use logarithms to find when this happens. Give your answer in hours and minutes, correct to the nearest minute.

1
(a) At $t = 0$ the power is $b^0 = 1$ , so the starting value gives $a$ directly.
$400 = a \times b^0 = a \quad \Rightarrow \quad a = 400$
2
Substitute the second point. $b$ must be positive for a growth model.
$900 = 400b^2 \quad \Rightarrow \quad b^2 = 2.25 \quad \Rightarrow \quad b = 1.5$
3
(b) Go across from $N = 2000$ to the curve, then down to the $t$ -axis. The reading is just under $4$ hours.
4
Set up the equation, divide by $400$ , then take logs of both sides.
$400 \times 1.5^{t} = 2000 \;\Rightarrow\; 1.5^{t} = 5 \;\Rightarrow\; t\log 1.5 = \log 5$
5
Divide. The answer agrees with the graph. ✓
$t = \frac{\log 5}{\log 1.5} = 3.969\ldots \approx 3.97 \text{ hours}$
6
(c) $N = 5000$ is beyond the graph, so logarithms are the only accurate method.
$400 \times 1.5^{t} = 5000 \;\Rightarrow\; 1.5^{t} = 12.5 \;\Rightarrow\; t = \frac{\log 12.5}{\log 1.5} = 6.2292\ldots$
7
Convert the decimal part of an hour to minutes.
$0.2292\ldots \times 60 = 13.75\ldots \approx 14 \text{ minutes}$

Answer

(a) $a = 400$ , $b = 1.5$ (b) about $4$ hours from the graph; $3.97$ hours (c) $6$ hours $14$ minutes

Examiner tip

Example 9ChallengeShow that: comparing two exponential models[8 marks]

When one population overtakes another

The populations of two towns are modelled by $P_A = 500 \times 1.08^{t}$ and $P_B = 900 \times 1.03^{t}$ , where $t$ is the time in years. The graphs of both models are shown.

(a) Use the graph to estimate the value of $t$ when the two populations are equal.

(b) Show that this value of $t$ satisfies $t = \dfrac{\log 1.8}{\log 1.08 - \log 1.03}$ .

(c) Find this value of $t$ correct to $3$ significant figures, and the populations at that time, correct to the nearest $10$ .

(d) Find the smallest whole number of years after which the population of town $A$ is greater than the population of town $B$ .

1
(a) The curves cross between $t = 12$ and $t = 13$ , a little under halfway, so $t \approx 12.4$ .
2
(b) Set the two models equal.
$500 \times 1.08^{t} = 900 \times 1.03^{t}$
3
Gather the powers on one side: divide both sides by $500$ and by $1.03^{t}$ .
$\frac{1.08^{t}}{1.03^{t}} = \frac{900}{500} = 1.8$
4
Take logs. The division law turns the left side into a difference.
$t\log 1.08 - t\log 1.03 = \log 1.8$
5
Factorise out $t$ and divide.
$t(\log 1.08 - \log 1.03) = \log 1.8 \;\Rightarrow\; t = \frac{\log 1.8}{\log 1.08 - \log 1.03}$
6
(c) Evaluate, keeping full accuracy. This agrees with the graph. ✓
$t = \frac{0.25527\ldots}{0.020586\ldots} = 12.39\ldots \approx 12.4$
7
Substitute the unrounded $t$ into either model — both give the same value.
$500 \times 1.08^{12.39\ldots} = 1298.4\ldots \approx 1300$
8
(d) Test the whole numbers either side of the crossing. After $12$ years $A$ is still smaller; after $13$ years it is larger.
$t = 12: \; 1259 < 1283, \qquad t = 13: \; 1360 > 1322$

Answer

(a) about $12.4$ (b) shown (c) $t = 12.4$ ; both populations about $1300$ (d) $13$ years

Examiner tip

Logarithms Worked ExamplesIGCSE Maths 0607

Logarithms from the definition

Combining logarithms

The power law

Solving an exponential equation

Solving a logarithmic equation

Exponential growth with logarithms

The graph of a logarithm as a reflection

Exponential model from a graph, solved with logarithms

When one population overtakes another

More IGCSE Maths 0607 worked examples

Functions

Number

Algebra

Coordinate Geometry

Geometry

Vectors and Transfomations

Mensuration

Trigonometry

Sets

Probability

Statistics

Logarithms Worked ExamplesIGCSE Maths 0607

Logarithms from the definition

Combining logarithms

The power law

Solving an exponential equation

Solving a logarithmic equation

Exponential growth with logarithms

The graph of a logarithm as a reflection

Exponential model from a graph, solved with logarithms

When one population overtakes another

More IGCSE Maths 0607 worked examples

Functions

Number

Algebra

Coordinate Geometry

Geometry

Vectors and Transfomations

Mensuration

Trigonometry

Sets

Probability

Statistics