A-Level Maths exponentials and logarithms: laws, equations and straight-line graphs

Where the topic sits in each specification

The key ideas

The three laws are log a + log b = log(ab), log a - log b = log(a/b), and k log a = log(a^k), all to the same base. They come from the index laws, so they work only for products, quotients and powers. There is no law for log(a + b).

e is the base for which the gradient of e^x equals e^x, and ln x means log base e. Because they are inverses, ln(e^x) = x and e^(ln x) = x for x > 0. To solve an equation with the unknown in a power, take ln (or log) of both sides, bring the power down, and collect the unknown. Equations such as e^(2x) - 5e^x + 6 = 0 are quadratics in disguise: put u = e^x.

If y = kx^n, then ln y = ln k + n ln x, so a graph of ln y against ln x is a straight line with gradient n and intercept ln k. If y = k a^x, then ln y = ln k + x ln a, so ln y against x is straight with gradient ln a. Exam questions give two points on the straight line, or a table of data, and ask for k and n or k and a.

Worked example 1: solve 3^(x + 1) = 4^(2x - 1), giving x to 3 significant figures

1. Take ln of both sides: (x + 1) ln 3 = (2x - 1) ln 4.
2. Expand: x ln 3 + ln 3 = 2x ln 4 - ln 4.
3. Collect x terms: ln 3 + ln 4 = 2x ln 4 - x ln 3 = x(2 ln 4 - ln 3).
4. So x = (ln 3 + ln 4)/(2 ln 4 - ln 3) = ln 12/ln(16/3) = 2.4849/1.6740.
5. Answer: x = 1.48 (3 s.f.). Check: 3^2.484 and 4^1.969 both come to about 15.3.

Worked example 2: reducing y = kx^n to a straight line

1. A graph of ln y against ln x is a straight line through (1.0, 2.5) and (3.0, 3.5). Find k and n.
2. Taking ln: ln y = ln k + n ln x, so the gradient is n and the intercept is ln k.
3. Gradient: n = (3.5 - 2.5)/(3.0 - 1.0) = 0.5.
4. Intercept: 2.5 = ln k + 0.5 × 1.0, so ln k = 2.0 and k = e^2 = 7.39 (3 s.f.).
5. Answer: y = 7.39 x^0.5.

Worked example 3: a growth model

1. A population is modelled by N = 500e^(0.04t), where t is in years. Find when the population first reaches 1500.
2. Set 500e^(0.04t) = 1500, so e^(0.04t) = 3.
3. Take ln: 0.04t = ln 3 = 1.0986.
4. So t = 1.0986/0.04 = 27.5 years (3 s.f.).

Common mistakes that cost marks

• Writing log(a + b) = log a + log b, or ln(x + 3) = ln x + ln 3.
• Dividing logs as if they were the log of a quotient: ln 12/ln 4 is not ln 3.
• Forgetting that ln x needs x > 0, so not rejecting solutions that make the argument negative.
• Reading the intercept of a log graph as k instead of ln k.
• Rounding too early in a chain of calculations, so the final answer is wrong at 3 s.f.
• On Cambridge papers, relying on change of base, which the 9709 syllabus excludes, rather than taking logs of both sides.

Exam technique and how a tutor helps

Show the line where you take logs of both sides and the line where the power comes down; those carry the method marks. Keep values exact or to at least 5 significant figures until the last step. For inequalities with logs, remember that dividing by a negative log, such as ln 0.5, reverses the inequality sign.

In one-to-one lessons a tutor checks first that the student really understands what a logarithm is, because many students learn the laws as symbols without meaning and then misapply them. Short daily questions on the laws, then mixed equations and log-graph questions from past papers, usually make this one of the most reliable sources of marks on the paper.

Self-check: can you do these?

• Solve e^(2x) - 5e^x + 6 = 0. (Answer: x = ln 2 or ln 3)
• Solve ln(2x - 1) = 3. (Answer: x = (e^3 + 1)/2, about 10.5)
• Solve 2^x = 7 to 3 s.f. (Answer: 2.81)
• Solve log_2(x + 6) - log_2 x = 2. (Answer: x = 2)
• If y = k(a^x), what do you plot to get a straight line, and what is the gradient? (Answer: ln y against x; gradient ln a)