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QLDMath MethodsSyllabus dot point

Topic 1: Exponential functions

Recall and apply the laws of indices to simplify expressions and solve equations involving rational and negative exponents

A focused answer to the QCE Math Methods Unit 2 dot point on the laws of indices. Lists the seven exponent laws, applies them to rational and negative powers, and works the QCAA-style equation a2x+1=ax3a^{2x+1} = a^{x-3} style problem used in IA1 and EA.

Generated by Claude Opus 4.87 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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Jump to a section
  1. What this dot point is asking
  2. The index laws
  3. Common manipulations
  4. Recognising base families
  5. Solving exponential equations (same-base method)
  6. Why the laws extend to all exponents
  7. How this appears in IA1 and EA

What this dot point is asking

QCAA wants you to apply the laws of indices with confidence, including with rational and negative exponents, and to use the laws to solve exponential equations where both sides can be rewritten with the same base.

The index laws

For any positive base aa and rationals mm, nn:

aman=am+na^m \cdot a^n = a^{m+n}

aman=amn\frac{a^m}{a^n} = a^{m-n}

(am)n=amn(a^m)^n = a^{mn}

(ab)n=anbn(ab)^n = a^n b^n

(ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

a0=1(a0)a^0 = 1 \quad (a \neq 0)

an=1ana^{-n} = \frac{1}{a^n}

Rational exponents: a1/n=ana^{1/n} = \sqrt[n]{a} and am/n=(an)ma^{m/n} = (\sqrt[n]{a})^m.

These laws come directly from the definition of repeated multiplication and extend smoothly to negative and rational powers.

Common manipulations

The recurring skill is to rewrite every term as a power of one base before combining.

Negative exponent
52=152=1255^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}, moving the power to the denominator.
Rational exponent
82/3=(83)2=22=48^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4, taking the cube root then squaring.
Same base products
2x23x+1=24x+12^x \cdot 2^{3x + 1} = 2^{4x + 1}, adding exponents.
Rebasing
9x+1=(32)x+1=32x+29^{x + 1} = (3^2)^{x + 1} = 3^{2x + 2}, which lets a base-99 term combine with base-33 terms.

These manipulations are also the gateway to differentiating and integrating power functions, where an expression like 1x\dfrac{1}{\sqrt x} must first be rewritten as x1/2x^{-1/2} before the power rule can apply.

Recognising base families

The same-base method depends on spotting that several numbers are powers of one base. The powers of 22 are 2,4,8,16,32,2, 4, 8, 16, 32, \ldots; the powers of 33 are 3,9,27,81,3, 9, 27, 81, \ldots; the powers of 55 are 5,25,125,5, 25, 125, \ldots. When a question mixes, say, 44 and 88 and 1616, every term can be written with base 22, after which the laws combine them and the exponents can be equated. Building familiarity with these families makes most Paper 1 index questions immediate, because the rewrite is the only hard step.

Solving exponential equations (same-base method)

If both sides of an equation can be rewritten with the same base, equate exponents and solve. This works when the bases are related by integer or rational powers (for example 4,8,164, 8, 16 all relate to base 22; 9,27,819, 27, 81 all relate to base 33). The justification is that the exponential function is one-to-one: if ap=aqa^p = a^q for a fixed base a>0a > 0, a1a \neq 1, then p=qp = q, so matching the powers is valid.

When the bases cannot be aligned (for example 2x=72^x = 7), logarithms are required (the next dot point covers this).

Why the laws extend to all exponents

The laws originate in repeated multiplication: a3a2=(aaa)(aa)=a5a^3 \cdot a^2 = (a \cdot a \cdot a)(a \cdot a) = a^5. To keep the rule am/an=amna^m / a^n = a^{m-n} working when m=nm = n forces a0=1a^0 = 1, and when m<nm < n it forces ak=1aka^{-k} = \tfrac{1}{a^k}. Likewise, requiring (a1/n)n=a1=a(a^{1/n})^n = a^{1} = a forces a1/n=ana^{1/n} = \sqrt[n]{a}. So the negative and fractional cases are not new rules but the only definitions that keep the original laws consistent, which is why they can be applied with confidence.

How this appears in IA1 and EA

IA1
Direct application questions: simplify a multi-term expression with mixed positive, negative and fractional exponents, or solve a same-base equation.
EA Paper 1
Multiple choice on index manipulations and short solve problems.
EA Paper 2
Used as the algebra step inside a larger problem (an exponential growth model, a calculus-of-exponentials computation).

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20223 marksPaper 1 (technique). Solve for xx: 23x+1=16x22^{3x + 1} = 16^{x - 2}.
Show worked answer →

Write both sides with base 22: 16=2416 = 2^4, so 16x2=24(x2)=24x816^{x - 2} = 2^{4(x - 2)} = 2^{4x - 8}.

Equate exponents: 3x+1=4x83x + 1 = 4x - 8, giving x=9x = 9.

Markers reward the common-base rewrite, equating exponents, and the linear solve.

QCAA 20234 marksPaper 1 (technique). (a) Simplify a5b2a2b\dfrac{a^5 b^{-2}}{a^2 b}, expressing the answer with positive indices. (b) Solve 9x=27x19^{x} = 27^{x - 1}.
Show worked answer →

(a) a5b2a2b=a52b21=a3b3=a3b3.\dfrac{a^5 b^{-2}}{a^2 b} = a^{5 - 2} b^{-2 - 1} = a^3 b^{-3} = \dfrac{a^3}{b^3}.

(b) Base 33: 9x=32x9^x = 3^{2x} and 27x1=33(x1)=33x327^{x - 1} = 3^{3(x - 1)} = 3^{3x - 3}. Equate exponents: 2x=3x32x = 3x - 3, so x=3x = 3.

Markers reward subtracting indices on division, expressing with positive indices, and the same-base solve.

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