IP AMaths Notes (Upper Sec, Year 3-4): 02) Logarithms and Exponentials

Logarithms invert exponentials. Switch comfortably between both forms to linearise growth models and solve equations involving powers.

Core facts

• Definition: \( \log_{a} b = c \) iff \( a^{c} = b \) for \( a > 0 \), \( a \neq 1 \), \( b > 0 \).
• Laws mirror index rules: \( \log_{a}(xy) = \log_{a} x + \log_{a} y \), \( \log_{a} \dfrac{x}{y} = \log_{a} x - \log_{a} y \).
• Power rule: \( \log_{a}(x^{k}) = k \log_{a} x \).
• Change of base: \( \log_{a} b = \dfrac{\log_{c} b}{\log_{c} a} \); choose \( c = 10 \) or \( c = e \) for calculator work.

Worked example 1 — Solve logarithmic equation

Solve \( \log_{2}(x + 3) + \log_{2}(x - 1) = 3 \).

1. Combine using product law: \( \log_{2}\bigl((x + 3)(x - 1)\bigr) = 3 \).
2. Convert to index form: \( (x + 3)(x - 1) = 2^{3} = 8 \).
3. Expand: \( x^{2} + 2x - 3 = 8 \) so \( x^{2} + 2x - 11 = 0 \).
4. Solve quadratic: \( x = \dfrac{-2 \pm \sqrt{4 + 44}}{2} = \dfrac{-2 \pm \sqrt{48}}{2} = -1 \pm 2\sqrt{3} \).
5. Domain check: require \( x > 1 \), so \( x = -1 - 2\sqrt{3} \) is extraneous. Valid solution: \( x = -1 + 2\sqrt{3} \).

Worked example 2 — Linear law

The model \( y = Ab^x \) can be linearised.

Given data points \( (x, y) = (1, 6.2) \) and \( (4, 20.5) \), show the straight-line form and estimate \( A \) and \( b \).

1. Take base-10 logs: \( \log_{10} y = \log_{10} A + x \log_{10} b \).
2. Treat \( Y = \log_{10} y \), \( C = \log_{10} A \), \( M = \log_{10} b \); now \( Y = C + M x \).
3. Substitute data: for \( x = 1 \), \( Y_1 = \log_{10} 6.2 = 0.7924 \) (4 s.f.). For \( x = 4 \), \( Y_2 = \log_{10} 20.5 = 1.3118 \).
4. Gradient: \( M = \dfrac{1.3118 - 0.7924}{4 - 1} = 0.1731 \).
5. Intercept: \( C = Y_1 - M = 0.7924 - 0.1731 = 0.6193 \).
6. Recover constants: \( b = 10^{M} = 10^{0.1731} = 1.49 \) (3 s.f.), \( A = 10^{C} = 10^{0.6193} = 4.17 \).

Try this

For \( 3\log_{5}(x - 1) = \log_{5}(9x + 1) \), solve for \( x \) and justify any rejected roots.