Everything here is mechanical once the laws are automatic. TMUA questions hide these skills inside bigger problems (solving \(4^x = 8^{2x-3}\), simplifying an integrand before integrating, or checking whether a surd expression is rational), so aim for speed and accuracy rather than depth.
Laws of indices (MM1.1)
- \(a^m \times a^n = a^{m+n}\), \(\quad a^m \div a^n = a^{m-n}\), \(\quad (a^m)^n = a^{mn}\), \(\quad (ab)^n = a^n b^n\)
- \(a^0 = 1\), \(\quad a^{-n} = \dfrac{1}{a^n}\), \(\quad a^{1/n} = \sqrt[n]{a}\), \(\quad a^{m/n} = \left(\sqrt[n]{a}\right)^m\)
- For fractional powers of fractions, take the root first: \(\left(\frac{27}{8}\right)^{-2/3} = \left(\frac{8}{27}\right)^{2/3} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
- Equations like \(4^{x+1} = 8^{2x-3}\): write everything as a power of one base (\(2^{2x+2} = 2^{6x-9}\)) and equate indices.
- 'Hidden quadratics': \(9^x - 4\cdot3^x + 3 = 0\) is \(y^2 - 4y + 3 = 0\) with \(y = 3^x\). Reject any root with \(y \leq 0\).
Surds (MM1.2, M2.12)
- \(\sqrt{ab} = \sqrt a\sqrt b\) and \(\sqrt{\frac ab} = \frac{\sqrt a}{\sqrt b}\) for \(a, b \geq 0\); but \(\sqrt{a + b} \neq \sqrt a + \sqrt b\).
- Simplify by extracting square factors: \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt3\).
- Rationalise with the conjugate: \(\dfrac{5}{3 + 2\sqrt5} = \dfrac{5(3 - 2\sqrt5)}{9 - 20} = \dfrac{-15 + 10\sqrt5}{11}\).
- \((a + \sqrt b)(a - \sqrt b) = a^2 - b\) is the difference of two squares; use it to compare or telescope surd sums, e.g. \(\frac{1}{\sqrt2 + 1} = \sqrt2 - 1\).
- \(\sqrt{x^2} = |x|\), never just \(x\): the symbol \(\sqrt{}\) always means the non-negative root (MM1.7).
Must-know (no formula booklet)
- \(a^{m/n} = (\sqrt[n]{a})^m\)
- \(a^{-n} = 1/a^n\)
- \(\sqrt{ab} = \sqrt a \sqrt b\)
- \((a+\sqrt b)(a-\sqrt b) = a^2 - b\)
- \(\sqrt{x^2} = |x|\)
Classic traps
- Do not add exponents when multiplying different bases: \(2^3 \times 3^2\) is not \(6^5\).
- \((x + y)^{1/2}\) cannot be split; only products and quotients split.
- When comparing surds like \(\sqrt2 + \sqrt7\) and \(2 + \sqrt5\), square them (both positive) instead of estimating.
Worked example
Q. Solve \(2^{x^2} = 16 \times 2^{-3x}\).
A. \(2^{x^2} = 2^{4-3x}\), so \(x^2 + 3x - 4 = 0\), \((x + 4)(x - 1) = 0\): \(x = -4\) or \(x = 1\).