Exponentials, Logarithms & Linearising Non-Linear Data
Exponential functions and logarithms appear across Pure Mathematics, Differential Equations, and Statistics. At A-Level, you must move fluently between exponential and logarithmic forms, solve disguised quadratic equations in e^x or a^x, and differentiate e^(kx) to find instantaneous rates of growth or decay. A guaranteed exam question tests your ability to reduce non-linear relationships, specifically power laws y = ax^n and exponential laws y = ab^x - into straight-line graphs by plotting log(y) against log(x) or log(y) against x.
- Laws of Logarithms and Change of Base
- For any base a > 0 (a ≠ 1) and strictly positive arguments x, y > 0, the equivalence is log_a(x) = n ⇔ a^n = x. The three core laws are: 1. Multiplication law: log_a(xy) = log_a(x) + log_a(y); 2. Division law: log_a(x / y) = log_a(x) - log_a(y); 3. Power law: log_a(x^k) = k log_a(x) (including log_a(1 / x) = -log_a(x)). Special values: log_a(a) = 1 and log_a(1) = 0. The change of base rule states log_a(b) = log_c(b) / log_c(a), so in particular log_a(b) = 1 / log_b(a).
- The Natural Exponential e^x and Natural Logarithm ln(x)
- Euler's number e ≈ 2.71828 is the unique base for which the gradient of y = e^x at every point is equal to its y-coordinate: d/dx(e^x) = e^x. More generally, by the chain rule, d/dx(A e^(kx)) = kA e^(kx). The natural logarithm y = ln(x) = log_e(x) is the inverse function of y = e^x, so its graph is the reflection of y = e^x in the line y = x. Consequently, e^(ln x) = x (for x > 0) and ln(e^x) = x (for all real x), with y = e^x having horizontal asymptote y = 0 and y = ln(x) having vertical asymptote x = 0.
- Disguised Quadratics in Exponential and Log Equations
- An equation such as 3^(2x) - 10(3^x) + 9 = 0 or e^x + 15e^(-x) = 8 is a disguised quadratic. Since 3^(2x) = (3^x)² and e^(-x) = 1 / e^x, substitute u = 3^x (or multiply through by e^x and let u = e^x) to form a quadratic in u. After factorising to find the roots for u, remember that u = a^x > 0 for all real x: any negative or zero root for u must be rejected as having no real solutions.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Laws of Logarithms and Change of Base
- For any base a > 0 (a ≠ 1) and strictly positive arguments x, y > 0, the equivalence is log_a(x) = n ⇔ a^n = x. The three core laws are: 1. Multiplication law: log_a(xy) = log_a(x) + log_a(y); 2. Division law: log_a(x / y) = log_a(x) - log_a(y); 3. Power law: log_a(x^k) = k log_a(x) (including log_a(1 / x) = -log_a(x)). Special values: log_a(a) = 1 and log_a(1) = 0. The change of base rule states log_a(b) = log_c(b) / log_c(a), so in particular log_a(b) = 1 / log_b(a).
- The Natural Exponential e^x and Natural Logarithm ln(x)
- Euler's number e ≈ 2.71828 is the unique base for which the gradient of y = e^x at every point is equal to its y-coordinate: d/dx(e^x) = e^x. More generally, by the chain rule, d/dx(A e^(kx)) = kA e^(kx). The natural logarithm y = ln(x) = log_e(x) is the inverse function of y = e^x, so its graph is the reflection of y = e^x in the line y = x. Consequently, e^(ln x) = x (for x > 0) and ln(e^x) = x (for all real x), with y = e^x having horizontal asymptote y = 0 and y = ln(x) having vertical asymptote x = 0.
- Disguised Quadratics in Exponential and Log Equations
- An equation such as 3^(2x) - 10(3^x) + 9 = 0 or e^x + 15e^(-x) = 8 is a disguised quadratic. Since 3^(2x) = (3^x)² and e^(-x) = 1 / e^x, substitute u = 3^x (or multiply through by e^x and let u = e^x) to form a quadratic in u. After factorising to find the roots for u, remember that u = a^x > 0 for all real x: any negative or zero root for u must be rejected as having no real solutions.
- Linearising Power Laws: y = ax^n
- Taking log₁₀ (or ln) of both sides of the power model y = ax^n gives: log₁₀(y) = log₁₀(ax^n) = log₁₀(a) + log₁₀(x^n) = n log₁₀(x) + log₁₀(a). Comparing this with the straight-line equation Y = mX + c shows that if you plot Y = log₁₀(y) on the vertical axis against X = log₁₀(x) on the horizontal axis, you get a straight line with gradient m = n and vertical intercept c = log₁₀(a) (so a = 10^c).
- Linearising Exponential Laws: y = ab^x (or y = Ae^(kx))
- Taking log₁₀ of both sides of the exponential model y = ab^x gives: log₁₀(y) = log₁₀(ab^x) = log₁₀(a) + log₁₀(b^x) = [log₁₀(b)]x + log₁₀(a). Comparing with Y = mX + c shows that plotting Y = log₁₀(y) on the vertical axis against X = x (not log x!) on the horizontal axis yields a straight line with gradient m = log₁₀(b) (so b = 10^m) and vertical intercept c = log₁₀(a) (so a = 10^c). Similarly, for y = Ae^(kx), plotting ln(y) against x gives gradient m = k and intercept c = ln(A).
- Exponential Growth, Decay, and Model Limitations
- In a continuous model N = A + B e^(-kt) (where k > 0), the initial value at t = 0 is N₀ = A + B (since e⁰ = 1). As t → ∞, e^(-kt) → 0, so N approaches a long-term limiting value (horizontal asymptote) of N → A. The instantaneous rate of change is dN/dt = -kB e^(-kt). In pure exponential growth P = P₀ e^(kt) (k > 0), the model eventually fails for large t because unlimited exponential growth is restricted in reality by finite resources, space, disease, or market saturation.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
When linearising data, how do the horizontal axes differ between testing y = ax^n and testing y = ab^x?
If a graph of log₁₀(y) against x has gradient -0.3 and vertical intercept 2, what is the equation for y in terms of x?
Why must you check your solutions in the original equation after solving a logarithmic equation like log₂(x) + log₂(x - 3) = 2?
What is the derivative with respect to t of P = 500 - 380e^(-0.05t)?
How do you solve 5^(2x - 1) = 3^(x + 2) for an exact answer in terms of logarithms?
What are the domain, range, and asymptote of f(x) = ln(2x - 6)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Solve the equation 2 log₃(x) - log₃(x - 2) = 2, showing every algebraic step and justifying why both or only one solution is valid. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Use the power law on the first term: log₃(x²) - log₃(x - 2) = 2 [1 mark]. 2. Use the division law to combine into a single logarithm: log₃[x² / (x - 2)] = 2 [1 mark]. 3. Convert from logarithmic to exponential form using 3² = 9: x² / (x - 2) = 9 [1 mark]. 4. Rearrange into a quadratic equation and factorise: x² = 9x - 18 → x² - 9x + 18 = 0 → (x - 3)(x - 6) = 0, giving x = 3 or x = 6 [1 mark]. 5. Check domain validity (require x > 0 and x - 2 > 0, i.e., x > 2): since both 3 > 2 and 6 > 2, both x = 3 and x = 6 are valid solutions [1 mark].
- Q2
Solve the equation 2e^x - 9 + 4e^(-x) = 0, giving your answers in exact logarithmic form. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Multiply every term by e^x (since e^x ≠ 0) to clear the negative exponent: 2(e^x)² - 9e^x + 4 = 0 [1 mark]. 2. Let u = e^x, transforming the equation into the quadratic 2u² - 9u + 4 = 0 [1 mark]. 3. Factorise the quadratic: (2u - 1)(u - 4) = 0 [1 mark]. 4. Solve for u: u = 1/2 or u = 4 (both are positive, which is required since e^x > 0 for all real x) [1 mark]. 5. Solve e^x = u using natural logarithms: e^x = 1/2 gives x = ln(1/2) = -ln(2), and e^x = 4 gives x = ln(4) = 2 ln(2) [1 mark].
- Q3
The mass M grams of a radioactive substance at time t years is modelled by the equation M = M₀ e^(-kt), where M₀ and k are positive constants. When A graph of ln(M) is plotted against t, the line passes through the points (0, 5.30) and (8, 3.70). Find the value of M₀ to the nearest integer, find the value of k, and calculate the half-life of the substance (the time taken for M to halve). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Take natural logarithms of M = M₀ e^(-kt): ln(M) = ln(M₀) - kt [1 mark]. 2. Identify the vertical intercept at t = 0: ln(M₀) = 5.30 → M₀ = e^(5.30) = 200.33. = 200 g (to the nearest integer) [1 mark]. 3. Find the gradient of the line between (0, 5.30) and (8, 3.70): -k = (3.70 - 5.30) ÷ (8 - 0) = -1.60 ÷ 8 = -0.20, so k = 0.20 yr⁻¹ [1 mark]. 4. Set M = (1/2)M₀ to find the half-life T: (1/2)M₀ = M₀ e^(-0.20 T) → e^(-0.20 T) = 0.5 [1 mark]. 5. Take natural logs and solve for T: -0.20 T = ln(0.5) = -ln(2) → T = ln(2) ÷ 0.20 = 3.47 years [1 mark].
- Q4
Two variables x and y are related by a power law of the form y = ax^n. When log₁₀(y) is plotted on the vertical axis against log₁₀(x) on the horizontal axis, a straight line is obtained that passes through the points (1.2, 2.9) and (3.2, 1.9). Find the exact values of n and a, and find the value of x when y = 50. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Linearise y = ax^n using base-10 logs: log₁₀(y) = n log₁₀(x) + log₁₀(a), where n is the gradient and log₁₀(a) is the vertical intercept [1 mark]. 2. Calculate the gradient n from the two points: n = (1.9 - 2.9) ÷ (3.2 - 1.2) = -1.0 ÷ 2.0 = -0.5 (or -1/2) [1 mark]. 3. Substitute (1.2, 2.9) and n = -0.5 into Y = nX + c: 2.9 = (-0.5)(1.2) + log₁₀(a) → log₁₀(a) = 2.9 + 0.6 = 3.5 [1 mark]. 4. Solve for a: a = 10^(3.5) = 10³ × √10 = 1000√10 (≈ 3162), so y = 10^(3.5) x^(-0.5) [1 mark]. 5. Substitute y = 50: 50 = 10^(3.5) x^(-0.5) → x^(0.5) = 10^(3.5) ÷ 50 = 63.2455. → x = (10^7) ÷ 2500 = 4000 [1 mark].
Common Exam Questions (4)
What is the difference between log_a(x + y) and log_a(x) + log_a(y)?
log_a(x) + log_a(y) equals log_a(xy) (the log of a product). There is NO logarithm law that simplifies log_a(x + y) (the log of a sum): never expand log_a(x + y) as log_a(x) + log_a(y)!
How do I solve an equation where x appears both in a base-2 log and a base-x log, like log₂(x) + 6 log_x(2) = 5?
Use the reciprocal change-of-base identity log_x(2) = 1 / log₂(x). Let u = log₂(x), which turns the equation into u + 6 / u = 5 → u² - 5u + 6 = 0 → u = 2 or u = 3, giving x = 2² = 4 or x = 2³ = 8.
Why do examiners ask for 'a limitation of the exponential model' at the end of modelling questions?
Pure exponential growth y = Ae^(kt) predicts that a population, company revenue, or infection count will grow to infinity as t → ∞. To get the mark, relate the limitation to the specific context: e.g., 'The model predicts the squirrel population grows without bound, which is unrealistic due to limited food supply and habitat.'
Can I use natural logs (ln) instead of log₁₀ when solving exponential equations like 3^x = 20?
Yes. Unless the question specifies a particular base (such as when a graph has axes labelled log₁₀ y), taking ln of both sides is standard practice across all A-Level Maths papers.
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