Exponentials and logarithms
Exponentials and logarithms tests the laws of indices and logs, the graphs of exponential and logarithmic functions, solving equations that mix exponentials, and modelling growth and decay. The natural exponential e and natural logarithm ln are central, and many questions require you to reduce a disguised exponential equation to a quadratic.
Laws and definitions
logₐ x is the power to which a must be raised to give x, so logₐ a = 1 and logₐ 1 = 0. The laws are log(xy) = log x + log y, log(x/y) = log x − log y and log(xⁿ) = n log x. Change of base uses logₐ b = log b / log a. Note that log(x + y) cannot be simplified.
Exponential and logarithmic graphs
y = aˣ passes through (0, 1) and has the x-axis as an asymptote; it increases for a > 1 and decreases for 0 < a < 1. y = ln x is defined only for x > 0, passes through (1, 0) and grows without bound, slowly. Because ln and e are inverse functions, e^(ln x) = x and ln(eˣ) = x.
Solving and modelling
To solve a^(f(x)) = b, take logs of both sides. Equations such as 3^(2x) − 5·3ˣ + 6 = 0 are quadratic in 3ˣ. In growth models y = Ae^(kt), k > 0 gives growth and k < 0 gives decay; taking logs of y = Abˣ gives a straight-line relationship, which is how data can be tested for exponential behaviour.
Worked example
Solve 3^(2x) − 5·3ˣ + 6 = 0.
Let y = 3ˣ. Then y² − 5y + 6 = 0, so (y − 2)(y − 3) = 0, giving y = 2 or y = 3. Hence 3ˣ = 2 or 3ˣ = 3, so x = log₃ 2 ≈ 0.631 or x = 1.
Traps that cost marks
- Writing log(a + b) = log a + log b, which is false.
- Taking logs of a negative or zero quantity, which is undefined.
- Forgetting that a substitution such as y = 3ˣ must be positive, discarding negative roots.
- Using log instead of ln in a derivative or integral involving e.