Exponentials and Logarithms
Exponentials and Logarithms
Exponentials and logarithms test the laws of logs, the special base e, solving equations where the unknown sits in an index, modelling growth and decay, and reducing a power or exponential law to linear form for graphical work. The paper expects confident switching between exponential and logarithmic form.
Laws and definitions
log a = x means a = 10^x for common logs, and ln a = x means a = e^x. Laws: log(xy) = log x + log y, log(x/y) = log x − log y, log xⁿ = n log x, log_a a = 1, log 1 = 0. Changing base: log_a b = log b / log a. Note e^ln x = x and ln e^x = x.
Solving index equations
For a^x = b take logs of both sides: x = log b / log a. For a quadratic in disguise, substitute u = a^x, solve the quadratic in u, then take logs of each positive root. Negative roots must be rejected because a^x > 0 for all real x.
Growth, decay and linearisation
If y = Ae^(kx), then ln y = ln A + kx, so plotting ln y against x gives gradient k and intercept ln A. If y = axⁿ, then ln y = ln a + n ln x, so the gradient of ln y against ln x is n. For decay models the constant k is negative and the half-life follows from e^(kt) = ½.
Worked example
Solve 2^(2x) − 5(2^x) + 4 = 0. Put u = 2^x, so u² − 5u + 4 = 0, giving (u − 1)(u − 4) = 0, hence u = 1 or u = 4. Since u = 2^x, 2^x = 1 gives x = 0, and 2^x = 4 = 2² gives x = 2. Both roots are positive values of u, so both are valid: x = 0 or x = 2. Substituting back confirms 1 − 5 + 4 = 0 and 16 − 20 + 4 = 0.
Traps that cost marks
- Writing log(x + y) = log x + log y, which is false.
- Keeping a negative root from the quadratic in u.
- Forgetting that the domain of ln x is x > 0 when stating a solution set.
- In linearisation, mixing up which quantity is the gradient and which is the intercept.
- Rounding intermediate logs and losing accuracy in the final index.