Logarithmic and exponential functions for IGCSE Mathematics - Additional: Methods and Worked Examples

Master logarithms and exponentials or lose easy marks

Logarithms and exponential functions appear on nearly every IGCSE Additional Mathematics paper. They're worth a significant chunk of marks, and the good news is this: the rules are few, the patterns are predictable, and once you nail the method, you'll pick up marks fast. The bad news? Most students lose marks not because the topic is hard, but because they muddle the laws or skip steps the examiner needs to see.

Here's what you need to do. Learn the three laws of logarithms cold. Understand how exponential and logarithmic forms convert back and forth. Know your graphs. Then practise solving equations until the process is automatic. That's it. No mystery, no magic.

The basics: exponential and logarithmic form

Every logarithm question boils down to one idea. If a^x = b, then log_a(b) = x. That's the conversion. Exponential form to logarithmic form and back again.

Get this drilled into your head:

• 2^3 = 8 means log_2(8) = 3
• 10^2 = 100 means log_10(100) = 2
• e^1 = e means ln(e) = 1

The base stays the base. The exponent becomes the answer. The result becomes the input to the log. If you can convert fluently in both directions, you've already got the foundation right.

The three laws of logarithms

Three laws. That's all. Every logarithm manipulation question on the IGCSE Additional Mathematics paper uses one or more of these.

Two extra results you need to know without thinking:

• log_a(a) = 1 always. The log of the base equals 1.
• log_a(1) = 0 always. Any base raised to the power 0 gives 1.

Applying the laws: a worked example

Question: Express 2 log_5(3) + log_5(4) - log_5(6) as a single logarithm.

Step 1: Use the power law on the first term. 2 log_5(3) = log_5(3^2) = log_5(9).

Step 2: Use the product law. log_5(9) + log_5(4) = log_5(9 x 4) = log_5(36).

Step 3: Use the quotient law. log_5(36) - log_5(6) = log_5(36/6) = log_5(6).

Answer: log_5(6).

Notice the order: power law first, then product or quotient. Always deal with coefficients before combining terms. Students who try to add first and apply the power law later get tangled.

Change of base

Your calculator has log (base 10) and ln (base e). It doesn't have log base 7 or log base 3. So when a question asks you to evaluate log_3(20), you need the change of base formula:

log_a(b) = log_c(b) / log_c(a)

Pick any base c you like. Base 10 and base e both work. On Paper 2 (calculator paper), just punch it in:

log_3(20) = log(20) / log(3) = 1.3010... / 0.4771... = 2.727 (3 significant figures)

On Paper 1 (non-calculator), change of base questions usually simplify to exact values. The examiner won't ask you to evaluate log_3(20) without a calculator. They'll give you something like log_9(27), which you can handle by converting both to powers of 3.

Graphs of exponential and logarithmic functions

The IGCSE syllabus expects you to sketch these graphs and identify their key features. Here's what you need to know, stripped to the essentials.

The exponential function y = e^x

• Passes through (0, 1) because e^0 = 1
• Always positive. The curve never touches or crosses the x-axis.
• Asymptote: y = 0 (the x-axis). The curve approaches it as x goes to negative infinity but never reaches it.
• Increases steeply as x increases
• For y = e^(kx), the curve is steeper when k is larger. For y = e^(-x), the curve is reflected in the y-axis.

The logarithmic function y = ln(x)

• Passes through (1, 0) because ln(1) = 0
• Also passes through (e, 1) because ln(e) = 1
• Defined only for x > 0. You cannot take the logarithm of zero or a negative number.
• Asymptote: x = 0 (the y-axis). The curve approaches it as x goes to 0 from the right.
• Increases slowly, getting flatter as x grows

The inverse relationship

y = e^x and y = ln(x) are inverse functions. On a graph, one is the reflection of the other in the line y = x. This means:

• e^(ln(x)) = x for all x > 0
• ln(e^x) = x for all real x

IGCSE examiners love to test this. If a question gives you y = e^(2x + 1) and asks for the inverse, you take the natural log of both sides: ln(y) = 2x + 1, then rearrange for x. Swap x and y at the end to get the inverse function.

Solving exponential equations

This is where the marks are. Most exam questions on this topic ask you to solve an equation involving an unknown exponent. The method is always the same.

Type 1: a^x = b (direct log approach)

Question: Solve 5^x = 40.

1. Take logarithms of both sides. log(5^x) = log(40).
2. Apply the power law. x log(5) = log(40).
3. Divide. x = log(40) / log(5).
4. Calculator: x = 1.6021... / 0.6990... = 2.292 (3 s.f.).

Four steps. Show all four. The examiner awards method marks for steps 1 and 2 even if your arithmetic goes wrong at the end.

Type 2: equations that reduce to a quadratic

Question: Solve e^(2x) - 5e^x + 6 = 0.

Spot the structure. Let u = e^x. Then e^(2x) = (e^x)^2 = u^2. The equation becomes u^2 - 5u + 6 = 0.

1. Factorise: (u - 2)(u - 3) = 0.
2. So u = 2 or u = 3.
3. Replace u: e^x = 2 or e^x = 3.
4. Take ln of both sides: x = ln(2) or x = ln(3).
5. If the paper wants exact answers, leave them as ln(2) and ln(3). If it wants decimals, x = 0.693 or x = 1.099 (3 s.f.).

Type 3: equations involving ln

Question: Solve ln(3x - 1) = 2.

1. Convert to exponential form: 3x - 1 = e^2.
2. Solve for x: 3x = e^2 + 1, so x = (e^2 + 1)/3.
3. Calculator: x = (7.389 + 1)/3 = 2.796 (3 s.f.).

Two marks for the conversion, one for the algebra. Don't overcomplicate it.

Common mistakes: a do/don't table

Exam strategy for logarithm questions

A few practical points that win marks:

• Always state which law you're using. Writing "using the product law" or "by the power law" costs you three seconds and earns you a method mark if your arithmetic slips.
• On Paper 1, leave answers exact. ln(2), log_5(6), 3/2. Don't approximate unless the question says "give your answer correct to 3 significant figures."
• On Paper 2, give 3 significant figures unless told otherwise. And show the full calculator display first, then round. Write x = 2.29203... = 2.29 (3 s.f.).
• Check your domain. If a question asks you to solve ln(2x - 5) = 3, your final answer for x must satisfy 2x - 5 > 0, meaning x > 2.5. If it doesn't, something went wrong.
• When graphing, label the asymptote and the intercepts. A sketch of y = e^x without "y = 0" marked as an asymptote and "(0, 1)" labelled will lose marks even if the shape is perfect.

Self-check questions

Test yourself. No peeking at the answers until you've written your full working.

1. Express 3 log_2(5) - log_2(10) + log_2(4) as a single logarithm.
2. Solve 7^x = 200, giving your answer to 3 significant figures.
3. Solve e^(2x) - 7e^x + 12 = 0, giving exact answers.
4. Given that log_a(3) = p and log_a(5) = q, express log_a(45) in terms of p and q.
5. Sketch the graph of y = ln(x + 2), showing the asymptote and any axis intercepts.

Answers

1. 3 log_2(5) = log_2(125). Then log_2(125) - log_2(10) = log_2(125/10) = log_2(12.5). Then log_2(12.5) + log_2(4) = log_2(50). Answer: log_2(50).
2. x log(7) = log(200). x = log(200)/log(7) = 2.3010.../0.8451... = 2.72 (3 s.f.).
3. Let u = e^x. u^2 - 7u + 12 = 0. (u - 3)(u - 4) = 0. u = 3 or u = 4. x = ln(3) or x = ln(4).
4. 45 = 9 x 5 = 3^2 x 5. log_a(45) = log_a(3^2) + log_a(5) = 2p + q.
5. Asymptote at x = -2 (vertical line). Crosses the y-axis at (0, ln 2). Crosses the x-axis at (-1, 0) because ln(-1 + 2) = ln(1) = 0. Curve shape: standard ln curve shifted 2 units left.

If you got all five right with full working, this topic is exam-ready. If you slipped on any, go back to the relevant section and drill the method again. Logarithms reward repetition. The students who score full marks aren't the ones who found it easy first time - they're the ones who practised the patterns until the steps became automatic.