O-Level Additional Mathematics · Secondary 3 and 4
Exponential and Logarithmic Functions
Exponential functions such as y = 2^x and y = e^x describe quantities that grow or decay by a constant factor, such as populations, savings and radioactive substances. Logarithms are the inverse operation: logₐ y is the power you raise a to in order to get y. In A-Math you use the laws of logarithms and the change of base rule, solve exponential and logarithmic equations, and sketch their graphs. The natural logarithm ln x, with base e, links directly to differentiation and integration later in the course.
What the syllabus expects
- Convert between index form a^x = y and logarithmic form x = logₐ y
- Use the laws of logarithms: the product, quotient and power laws
- Use the change of base rule to evaluate or simplify logarithms with different bases
- Solve exponential equations, including those that become quadratic in a^x, such as 4^x - 3(2^x) - 4 = 0
- Solve logarithmic equations and reject any solution that makes a logarithm undefined
- Sketch the graphs of y = a^x, y = e^x, y = logₐ x and y = ln x and state their key features
- Use exponential and logarithmic models in simple problems on growth and decay
Index form and logarithmic form
If a^x = y, where a > 0 and a ≠ 1, then x = logₐ y. So 2⁵ = 32 means log₂ 32 = 5, and 10³ = 1000 means lg 1000 = 3, where lg means log₁₀. The natural logarithm ln x means the logarithm to base e, where e ≈ 2.718. Two values are worth remembering: logₐ 1 = 0 because a⁰ = 1, and logₐ a = 1 because a¹ = a. You can only take the logarithm of a positive number, so logₐ x is undefined when x ≤ 0. This is why you must check the solutions of logarithmic equations at the end. Converting between the two forms is often the first step in solving an equation, for example log₃ x = 4 gives x = 3⁴ = 81.
Laws of logarithms and change of base
For positive M and N: logₐ(MN) = logₐ M + logₐ N, logₐ(M/N) = logₐ M - logₐ N, and logₐ(M^k) = k logₐ M. These come straight from the laws of indices. There is no law for logₐ(M + N), so it cannot be split. The change of base rule is logₐ b = (log b)/(log a), where both logarithms on the right use the same base, usually 10 or e. For example, log₄ 8 = (lg 8)/(lg 4) = (3 lg 2)/(2 lg 2) = 3/2. A special case is that log₂ 9 = 1/(log₉ 2). To solve log₂ x + log₂(x - 2) = 3, combine the logs to get log₂(x(x - 2)) = 3, so x² - 2x = 8, which gives x = 4 or x = -2. Since log₂(-2) is undefined, x = 4 only.
Solving exponential equations
If both sides can be written as powers of the same base, compare indices: 8^x = 4^(x+1) gives 2^(3x) = 2^(2x+2), so 3x = 2x + 2 and x = 2. Otherwise, take logarithms of both sides: 3^x = 20 gives x lg 3 = lg 20, so x = (lg 20)/(lg 3) = 2.73 to 3 significant figures. Some equations are quadratics in disguise. In 4^x - 3(2^x) - 4 = 0, note that 4^x = (2^x)². Let u = 2^x, so u² - 3u - 4 = 0, which factorises as (u - 4)(u + 1) = 0. Then 2^x = 4 gives x = 2, while 2^x = -1 has no solution because 2^x is always positive.
Graphs and models
The graph of y = a^x with a > 1 passes through (0, 1), increases, and approaches the x-axis as x becomes very negative, so y = 0 is an asymptote. The graph of y = e^(-x) decreases instead, which models decay. The graph of y = logₐ x with a > 1 is the reflection of y = a^x in the line y = x: it passes through (1, 0), is defined only for x > 0, and has the y-axis as an asymptote. In a model such as N = 500e^(0.2t), the value when t = 0 is 500. To find when N reaches 2000, solve e^(0.2t) = 4, so 0.2t = ln 4 and t = 5 ln 4 ≈ 6.93.
Worked example
Solve the equation 2^(2x+1) - 5(2^x) + 2 = 0.
- Use the index laws: 2^(2x+1) = 2 × 2^(2x) = 2(2^x)².
- Let u = 2^x. The equation becomes 2u² - 5u + 2 = 0.
- Factorise: (2u - 1)(u - 2) = 0, so u = 1/2 or u = 2.
- 2^x = 1/2 = 2^(-1) gives x = -1.
- 2^x = 2 = 2¹ gives x = 1.
- Check x = 1: 2³ - 5(2) + 2 = 8 - 10 + 2 = 0. Check x = -1: 2⁻¹ - 5(1/2) + 2 = 1/2 - 5/2 + 2 = 0.
Answer: x = -1 or x = 1
Common mistakes
- Splitting log(M + N) into log M + log N. The product law is for log(MN). A sum inside a logarithm cannot be split.
- Writing (lg 20)/(lg 3) as lg(20/3) or as lg 20 - lg 3. A quotient of two logarithms is not the logarithm of a quotient.
- Keeping a solution that makes a logarithm undefined. Substitute each answer into every log term of the original equation and reject any that give the log of zero or of a negative number.
- Treating 2^(2x+1) as (2^x)² + 1. Use the index laws carefully: 2^(2x+1) = 2 × (2^x)².
- Accepting 2^x = -1 as giving a value of x. An exponential a^x is always positive, so this has no real solution.
- Writing logₐ(M^k) as (logₐ M)^k. The power law brings the index down as a multiplier, so logₐ(M^k) = k logₐ M.
Exam tips
- Look for a hidden quadratic whenever you see a^(2x) or a^(2x+k) together with a^x. Substitute u = a^x and remember that u must be positive.
- When logarithms have different bases, convert everything to one base with the change of base rule before combining. Choosing the smaller base, such as 2 instead of 4, usually keeps the numbers whole.
- Give final answers from logarithms to 3 significant figures unless told otherwise, but keep full calculator values in the intermediate steps.
Practice questions
1. Without using a calculator, evaluate log₂ 48 - log₂ 3.
Show answer
4. By the quotient law, log₂ 48 - log₂ 3 = log₂(48/3) = log₂ 16 = 4, since 2⁴ = 16.
2. Solve log₃(x + 6) + log₃ x = 3.
Show answer
x = 3. Combining gives log₃(x(x + 6)) = 3, so x² + 6x = 27 and x² + 6x - 27 = 0, which is (x + 9)(x - 3) = 0. Reject x = -9 because log₃(-9) is undefined. Check x = 3: log₃ 9 + log₃ 3 = 2 + 1 = 3.
3. Solve log₄ x = log₂(x - 6).
Show answer
x = 9. By change of base, log₄ x = (log₂ x)/(log₂ 4) = (log₂ x)/2. So log₂ x = 2 log₂(x - 6) = log₂(x - 6)², giving x = (x - 6)², so x² - 13x + 36 = 0 and (x - 4)(x - 9) = 0. Reject x = 4 because log₂(-2) is undefined. Check x = 9: log₄ 9 = log₂ 3 and log₂(9 - 6) = log₂ 3.
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