← Unit 2: Calculus

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 $a^{2x+1} = a^{x-3}$ style problem used in IA1 and EA.

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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:

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aβˆ’n=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

Negative exponent. 5βˆ’2=1/255^{-2} = 1/25.

Rational exponent. 82/3=(83)2=22=48^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4.

Same base. 2xβ‹…23x+1=24x+12^x \cdot 2^{3x+1} = 2^{4x+1}.

Different bases, rewritten. 9x+1=(32)x+1=32x+29^{x+1} = (3^2)^{x+1} = 3^{2x+2}, so 9x+1=3β‹…32x+19^{x+1} = 3 \cdot 3^{2x+1}.

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).

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

Worked example

Simplify (2x)3β‹…xβˆ’24x1/2\dfrac{(2x)^3 \cdot x^{-2}}{4x^{1/2}}.

Numerator: (2x)3=8x3(2x)^3 = 8x^3. So (2x)3β‹…xβˆ’2=8x3βˆ’2=8x(2x)^3 \cdot x^{-2} = 8 x^{3-2} = 8 x.

Divide: 8x4x1/2=2x1βˆ’1/2=2x1/2=2x\dfrac{8x}{4 x^{1/2}} = 2 x^{1 - 1/2} = 2 x^{1/2} = 2\sqrt{x}.

Common traps

Adding exponents on different bases. 23β‹…322^3 \cdot 3^2 is not 656^5. Index laws only work when bases match.

Misapplying the negative exponent. βˆ’24-2^4 is βˆ’16-16, not 1616. The negative is outside the power. (βˆ’2)4=16(-2)^4 = 16.

Forgetting to align bases. 4x=324^x = 32 becomes 22x=252^{2x} = 2^5, so x=5/2x = 5/2.

Treating 000^0 as 11. QCAA does not test 000^0. Avoid it; if it appears, treat it as undefined.

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).

In one sentence

Indices obey seven laws (aman=am+na^m a^n = a^{m+n}, am/an=amβˆ’na^m / a^n = a^{m-n}, (am)n=amn(a^m)^n = a^{mn}, product and quotient over different bases, a0=1a^0 = 1, aβˆ’n=1/ana^{-n} = 1/a^n, rational exponents), and any exponential equation whose two sides can be written with the same base reduces to equating the exponents.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC3 marksSolve for $x$: $2^{3x+1} = 16^{x-2}$.
Show worked answer β†’

Express both sides with the same base. 16=2416 = 2^4, so 16xβˆ’2=24(xβˆ’2)=24xβˆ’816^{x-2} = 2^{4(x-2)} = 2^{4x-8}.

Equate exponents: 3x+1=4xβˆ’83x + 1 = 4x - 8.

x=9x = 9.

Markers reward expressing both sides with base 22, equating exponents, and a clean linear solve.

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