Core Algebraic Manipulation & Growth Models

Indices, Surd Conjugates & Logarithmic Laws

Mastering exponential transformations, surd rationalization using difference of two squares $(a+\sqrt{b})(a-\sqrt{b})$, the logarithm change-of-base theorem, and solving non-linear exponential growth and decay models.

1. Core Formulae & Mathematical Identities

Laws of Indices

$a^m \cdot a^n = a^{m+n}$
$a^m \div a^n = a^{m-n}$
$(a^m)^n = a^{mn}$
$a^{-n} = \frac{1}{a^n} \quad a^{m/n} = \sqrt[n]{a^m}$

For quadratic indices like $2^{2x} - 5(2^x) + 4 = 0$, substitute $u = 2^x$.

Surds & Conjugates

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
$\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$
$\frac{1}{a + \sqrt{b}} \times \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}$

Always rationalize the denominator so no radical remains underneath.

Laws of Logarithms

$\log_a(xy) = \log_a x + \log_a y$
$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$
$\log_a(x^k) = k \log_a x$
$\log_a b = \frac{\log_c b}{\log_c a} = \frac{1}{\log_b a}$

Check for extraneous roots: argument of logarithm must be strictly positive ($x > 0$).

2. Progressive SPM Worked Examples with Marking Schemes

Example 1 • Rationalizing Surds [4 Marks]

Express $\frac{3 + \sqrt{5}}{2\sqrt{5} - 3}$ in the form $p + q\sqrt{5}$, where $p$ and $q$ are rational numbers.

Step-by-Step Marking Solution
Multiply numerator and denominator by the conjugate of $(2\sqrt{5} - 3)$, which is $(2\sqrt{5} + 3)$:
$\frac{3 + \sqrt{5}}{2\sqrt{5} - 3} \times \frac{2\sqrt{5} + 3}{2\sqrt{5} + 3}$ [K1]
Expand the numerator:
$(3 + \sqrt{5})(2\sqrt{5} + 3) = 6\sqrt{5} + 9 + 2(\sqrt{5})^2 + 3\sqrt{5} = 9 + 2(5) + 9\sqrt{5} = 19 + 9\sqrt{5}$ [K1]
Expand the denominator using $(a - b)(a + b) = a^2 - b^2$:
$(2\sqrt{5})^2 - (3)^2 = 4(5) - 9 = 20 - 9 = 11$
$\frac{19 + 9\sqrt{5}}{11} = \frac{19}{11} + \frac{9}{11}\sqrt{5}$
$\therefore p = \frac{19}{11}, \quad q = \frac{9}{11}$ [N1, N1]
Example 2 • Logarithmic Change of Base [5 Marks]

Solve the logarithmic equation: $$\log_2 x - 4\log_x 2 = 3$$

Step-by-Step Marking Solution
Apply the reciprocal change of base rule: $\log_x 2 = \frac{1}{\log_2 x}$ [P1]
Substitute into the equation:
$\log_2 x - \frac{4}{\log_2 x} = 3$
Let $u = \log_2 x$:
$u - \frac{4}{u} = 3 \implies u^2 - 4 = 3u \implies u^2 - 3u - 4 = 0$ [K1]
$(u - 4)(u + 1) = 0 \implies u = 4 \quad \text{or} \quad u = -1$
Convert back to $x$:
• $\log_2 x = 4 \implies x = 2^4 = 16$ [N1]
• $\log_2 x = -1 \implies x = 2^{-1} = \frac{1}{2}$ [N1]
Since both $x = 16$ and $x = \frac{1}{2}$ are positive and $\neq 1$, both solutions are valid.
Example 3 • KBAT Nuclear Medicine Isotope Decay [5 Marks]

In radiotherapy, Technetium-99m has a radioactive decay model given by: $$M(t) = M_0 e^{-0.1155 t}$$ where $M_0$ is the initial mass in milligrams and $t$ is the elapsed time in hours.

(a) Find the half-life of Technetium-99m (the time required for the mass to decay to half of its initial value). [2 marks]
(b) If an initial dose of $50\text{ mg}$ is prepared at 08:00, determine the time when the remaining mass drops to $5\text{ mg}$. [3 marks]
(a) Half-life: $M(t) = 0.5 M_0 \implies M_0 e^{-0.1155 t} = 0.5 M_0 \implies e^{-0.1155 t} = 0.5$
Take natural logarithm $\ln$: $-0.1155 t = \ln(0.5) \approx -0.69315$
$t = \frac{-0.69315}{-0.1155} \approx 6.00\text{ hours}$.
[K1] $e^{-0.1155t} = 0.5$
[N1] $6.00\text{ hours}$
(b) When $M(t) = 5\text{ mg}$ with $M_0 = 50$:
$50 e^{-0.1155 t} = 5 \implies e^{-0.1155 t} = \frac{5}{50} = 0.1$ [K1]
$-0.1155 t = \ln(0.1) \approx -2.30259$
$t = \frac{-2.30259}{-0.1155} \approx 19.936\text{ hours} \approx 19\text{ hours } 56\text{ minutes}$. [K1]
Elapsed time $= 19\text{ hours } 56\text{ minutes}$ after 08:00 $\implies 03:56$ the following morning. [N1] 03:56