Index and logarithm laws, factorisation techniques, solving polynomial equations, and the relationship between exponential and logarithmic functions

What this dot point is asking

QCAA wants Year 11 students to be fluent with index and logarithm laws, factorise polynomials by a range of techniques, and solve linear, quadratic, polynomial, exponential and logarithmic equations. These manipulation skills are the algebraic toolkit relied on throughout Methods, especially when calculus questions reduce to solving an equation.

Index laws

The index laws govern multiplication, division and powers of expressions sharing a base:

The special and negative cases follow from these:

Products and quotients distribute over a power:

A fluent simplification almost always rewrites every term with a common base before applying these laws, since the laws only combine powers of the same base.

Logarithm laws

$\log_a(mn) = \log_a m + \log_a n$.

$\log_a(m/n) = \log_a m - \log_a n$.

$\log_a(m^n) = n \log_a m$.

$\log_a(1) = 0$. $\log_a(a) = 1$.

Change of base: $\log_a x = \log_b x / \log_b a$.

Inverse: $\log_a(a^x) = x$, $a^{\log_a x} = x$.

Logarithms in detail

A logarithm answers "to what power must the base be raised?" so $\log_a x = y$ means $a^y = x$. The change-of-base rule $\log_a x = \dfrac{\log_b x}{\log_b a}$ lets a calculator (which has only base $10$ and base $e$) evaluate any logarithm, and it is also the route to solving an exponential equation whose two sides cannot be written with the same base: take a logarithm of both sides and the unknown exponent comes down as a multiplier. The laws for products, quotients and powers mirror the index laws, because a logarithm is the inverse operation of raising to a power.

Factorisation

Factorisation rewrites a polynomial as a product, which is the key to solving equations via the null factor law (if a product is zero, at least one factor is zero).

Common factor

$6x^3 - 9x^2 = 3x^2(2x - 3)$.

Grouping

$x^3 + 2x^2 - x - 2 = x^2(x + 2) - (x + 2) = (x + 2)(x^2 - 1)$.

Quadratic

$x^2 + 5x + 6 = (x + 2)(x + 3)$.

Quadratic formula

$x = (-b \pm \sqrt{b^2 - 4ac})/(2a)$.

Difference of squares

$a^2 - b^2 = (a-b)(a+b)$.

Sum/difference of cubes

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Solving equations

Linear

Single-step manipulation.

Quadratic

Factor first, use null factor law. Or quadratic formula.

Polynomial

Factor where possible. Find rational roots first; polynomial division for higher degree.

Exponential

Bring to common base, equate exponents. Otherwise take logs.

Logarithmic

Combine logs using laws; convert to exponential form. Always check domain.

The exponential-logarithm relationship

The logarithm is the inverse of the exponential: $y = a^x$ and $y = \log_a x$ undo each other, which is why $\log_a(a^x) = x$ and $a^{\log_a x} = x$. This relationship is the reason logarithms solve exponential equations: taking a logarithm of both sides brings the unknown exponent down as a coefficient. It also explains the graphs, which are reflections of each other in the line $y = x$, with the exponential's horizontal asymptote $y = 0$ becoming the logarithm's vertical asymptote $x = 0$.

Hidden quadratics

Many equations that look exponential are quadratics in disguise. An equation such as $2^{2x} - 5(2^x) + 4 = 0$ becomes a standard quadratic under the substitution $u = 2^x$, because $2^{2x} = (2^x)^2 = u^2$. The same trick handles equations in $e^{2x}$ and $e^x$, or in $x$ and $\sqrt{x}$. Spotting that one expression is the square of another is the key, after which familiar factoring finishes the job.