The Fundamental Laws of Indices
- The Power Concept: For base $a$ and exponent $n \in \mathbb{Z}^+$: $$a^n = \underbrace{a \times a \times \dots \times a}_{n \text{ factors}}$$ Critical Sign Distinction: $(-3)^2 = (-3) \times (-3) = +9$, whereas $-3^2 = -(3 \times 3) = -9$. Order of operations applies exponentiation before unary negation.
- Law 1 (Product Law): When multiplying terms with identical bases, add their exponents: $$a^m \times a^n = a^{m+n}$$
- Law 2 (Quotient Law): When dividing terms with identical bases, subtract the denominator exponent: $$\frac{a^m}{a^n} = a^{m-n} \quad (a \ne 0)$$
- Law 3 (Power of a Power): Multiply exponents when raising a power to another power: $$(a^m)^n = a^{m \cdot n}$$
- Law 4 (Power of a Product & Quotient): Exponents distribute across multiplication and division: $$(ab)^n = a^n b^n, \quad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (b \ne 0)$$
Example 1 · Criterion A
Simplifying Multi-Variable Index Expressions
Simplify the algebraic expression, expressing your final answer with positive indices only: $$\frac{(2x^3 y^{-2})^4 \times (3x^{-4} y^5)^2}{6x^{-5} y^3}$$
- Step 1: Distribute powers across brackets: $$(2x^3 y^{-2})^4 = 2^4 \cdot x^{3 \times 4} \cdot y^{(-2) \times 4} = 16 x^{12} y^{-8}$$ $$(3x^{-4} y^5)^2 = 3^2 \cdot x^{(-4) \times 2} \cdot y^{5 \times 2} = 9 x^{-8} y^{10}$$
- Step 2: Multiply the numerator terms: $$\text{Numerator} = (16 \times 9) \cdot x^{12 + (-8)} \cdot y^{-8 + 10} = 144 x^4 y^2$$
- Step 3: Divide by denominator: $$\frac{144 x^4 y^2}{6 x^{-5} y^3} = \left(\frac{144}{6}\right) \cdot x^{4 - (-5)} \cdot y^{2 - 3} = 24 x^9 y^{-1}$$
- Step 4: Express with positive indices: $$24 x^9 y^{-1} = \frac{24x^9}{y}$$
Answer: $\frac{24x^9}{y}$.
Zero and Negative Integer Exponents (Logical Proofs)
- Proof of the Zero Exponent Law ($a^0 = 1$): Using the quotient law on equal powers: $$\frac{a^n}{a^n} = a^{n - n} = a^0 \quad \text{for any } a \ne 0$$ Any non-zero quantity divided by itself equals $1$: $$\frac{a^n}{a^n} = 1 \implies a^0 = 1 \quad (a \ne 0)$$ Note: $0^0$ is mathematically indeterminate.
- Proof of the Negative Exponent Law ($a^{-n} = \frac{1}{a^n}$): Applying the quotient law with $m = 0$: $$\frac{1}{a^n} = \frac{a^0}{a^n} = a^{0 - n} = a^{-n}$$ Therefore, a negative exponent represents the multiplicative inverse (reciprocal): $$a^{-n} = \frac{1}{a^n}, \quad \frac{1}{a^{-n}} = a^n$$
- Fractional Bases with Negative Exponents: $$\left(\frac{a}{b}\right)^{-n} = \frac{1}{\left(\frac{a}{b}\right)^n} = \frac{1}{\frac{a^n}{b^n}} = \frac{b^n}{a^n} = \left(\frac{b}{a}\right)^n$$ Inverting the base fraction flips the sign of the exponent.
Example 2 · Criterion A & C
Evaluating Negative Powers and Fractions
Evaluate each expression as an exact integer or fraction:
(a) $5^{-3}$ (b) $\left(\frac{3}{4}\right)^{-2}$ (c) $\frac{2^{-1} + 3^{-1}}{2^{-1} - 3^{-1}}$
- (a) Simple reciprocal: $$5^{-3} = \frac{1}{5^3} = \frac{1}{125}$$
- (b) Inverted fraction: $$\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^2 = \frac{16}{9} = 1\frac{7}{9}$$
- (c) Compound negative powers: $$2^{-1} = \frac{1}{2}, \quad 3^{-1} = \frac{1}{3}$$ $$\text{Numerator} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6}$$ $$\text{Denominator} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$$ $$\frac{\frac{5}{6}}{\frac{1}{6}} = \frac{5}{6} \times \frac{6}{1} = 5$$
Answers: (a) $\frac{1}{125}$; (b) $\frac{16}{9}$; (c) $5$.
Solving Exponential Equations with Common Bases
- Base Equivalence Principle: $$\text{If } a^{f(x)} = a^{g(x)} \quad \text{and } a \notin \{-1, 0, 1\}, \quad \text{then } f(x) = g(x)$$
- General Solving Strategy:
• Step 1: Identify a common prime base (e.g., $4, 8, 16, 32 \to 2$; $9, 27, 81 \to 3$; $25, 125 \to 5$).
• Step 2: Express both sides as single powers of that base using $(a^m)^n = a^{mn}$.
• Step 3: Equate exponents and solve the resulting linear or quadratic equation. - Factoring Exponential Equations: When powers with sums appear (e.g., $2^{x+2} + 2^x = 40$), factor out the common power $2^x$: $$2^x(2^2 + 1) = 40 \implies 2^x(5) = 40 \implies 2^x = 8 = 2^3 \implies x = 3$$
Example 3 · Criterion A
Solving Base Equivalence Equations
Solve for $x \in \mathbb{R}$:
(a) $9^{2x - 1} = 27^{x + 2}$
(b) $4^{x+1} - 4^{x-1} = 60$
- (a) Common base $3$: $$9 = 3^2, \quad 27 = 3^3$$ $$(3^2)^{2x - 1} = (3^3)^{x + 2} \implies 3^{2(2x - 1)} = 3^{3(x + 2)}$$ $$3^{4x - 2} = 3^{3x + 6}$$ Equate exponents: $$4x - 2 = 3x + 6 \implies 4x - 3x = 6 + 2 \implies x = 8$$
- (b) Factoring common factor $4^{x-1}$ or $4^x$: $$4 \cdot 4^x - \frac{1}{4} \cdot 4^x = 60$$ $$4^x\left(4 - \frac{1}{4}\right) = 60 \implies 4^x\left(\frac{15}{4}\right) = 60$$ $$4^x = 60 \times \frac{4}{15} = 4 \times 4 = 16$$ $$4^x = 4^2 \implies x = 2$$
Answers: (a) $x = 8$; (b) $x = 2$.
Simplify each expression, expressing your answer with positive exponents:
(a) $p^5 \times p^{-8}$ (b) $\frac{q^{-3}}{q^4}$ (c) $(r^{-2})^{-5}$
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(a) $p^{5 + (-8)} = p^{-3} = \frac{1}{p^3}$ [A1]
(b) $q^{-3 - 4} = q^{-7} = \frac{1}{q^7}$ [A1]
(c) $r^{(-2) \times (-5)} = r^{10}$ [A1]
Evaluate each numerical expression without a calculator:
(a) $7^0 + 4^{-2}$ (b) $\left(-\frac{2}{3}\right)^{-3}$ (c) $\frac{3^{-2}}{6^{-2}}$
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(a) $7^0 = 1$; $4^{-2} = \frac{1}{16} \implies 1 + \frac{1}{16} = \frac{17}{16}$ [A1]
(b) $\left(-\frac{3}{2}\right)^3 = -\frac{27}{8}$ [A1]
(c) $\left(\frac{3}{6}\right)^{-2} = \left(\frac{1}{2}\right)^{-2} = 2^2 = 4$ [A1]
Simplify the algebraic fraction completely: $$\frac{(3a^2 b^{-3})^3 \times (2a^{-1} b^4)^2}{18 a^5 b^{-2}}$$
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Step 1: Expand powers [2 marks]
$(3a^2 b^{-3})^3 = 27 a^6 b^{-9}$ [A1]
$(2a^{-1} b^4)^2 = 4 a^{-2} b^8$ [A1]
Step 2: Combine numerator and divide [2 marks]
Numerator $= (27 \times 4) a^{6 - 2} b^{-9 + 8} = 108 a^4 b^{-1}$
Division: $\frac{108 a^4 b^{-1}}{18 a^5 b^{-2}} = \left(\frac{108}{18}\right) a^{4 - 5} b^{-1 - (-2)} = 6 a^{-1} b^1 = \frac{6b}{a}$ [A2]
Solve the exponential equation for $x$: $$2^{3x - 5} = 128$$
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Step 1: Write with common base $2$ [1 mark]
$128 = 2^7$ [M1]
Step 2: Equate exponents and solve [2 marks]
$3x - 5 = 7$ [M1]
$3x = 12 \implies x = 4$ [A1]
Solve the exponential equation for $x$: $$25^{x+1} = 125^{2x-3}$$
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Step 1: Express both sides in base $5$ [2 marks]
$25 = 5^2 \implies (5^2)^{x+1} = 5^{2x + 2}$ [A1]
$125 = 5^3 \implies (5^3)^{2x-3} = 5^{6x - 9}$ [A1]
Step 2: Equate exponents [2 marks]
$2x + 2 = 6x - 9$ [M1]
$4x = 11 \implies x = \frac{11}{4} = 2.75$ [A1]
Investigate the last digit (units digit) of powers of $7$:
• $7^1 = 7$, $7^2 = 49$, $7^3 = 343$, $7^4 = 2401$, $7^5 = 16807$.
(a) State the repeating pattern and period length of the units digit of $7^n$. [2 marks]
(b) Predict the units digit of $7^{82}$. Justify your reasoning using modular division. [2 marks]
(c) State the units digit of $7^{2025} + 3$. [1 mark]
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(a) Pattern and cycle length: [2 marks]
The units digits cycle as $\{7, 9, 3, 1\}$ with period length $T = 4$. [A2]
(b) Units digit of $7^{82}$: [2 marks]
Divide exponent by period: $82 \div 4 = 20\text{ remainder } 2$ [M1].
The remainder of $2$ corresponds to the 2nd digit in the cycle $\implies \mathbf{9}$. [A1]
(c) Units digit of $7^{2025} + 3$: [1 mark]
$2025 \div 4 = 506\text{ remainder } 1$. Last digit of $7^{2025}$ is $7$.
Units digit of $7 + 3 = 10 \implies \mathbf{0}$. [A1]
Show that for all positive integers $n$, the expression $\frac{3^{n+3} - 3^{n+1}}{2 \cdot 3^n}$ is an integer constant independent of $n$. Find its value.
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Step 1: Factor out $3^n$ from numerator [2 marks]
$3^{n+3} = 3^n \cdot 3^3 = 27 \cdot 3^n$
$3^{n+1} = 3^n \cdot 3^1 = 3 \cdot 3^n$
$\text{Numerator} = 3^n(27 - 3) = 24 \cdot 3^n$ [A2]
Step 2: Simplify with denominator [2 marks]
$$\frac{24 \cdot 3^n}{2 \cdot 3^n} = \frac{24}{2} = 12$$
Since $12$ contains no variable $n$, the expression is constant for all $n \in \mathbb{Z}^+$. [A2]
Solve the equation for $x$: $$3^{2x} - 10 \cdot 3^x + 9 = 0$$
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Step 1: Quadratic substitution [1 mark]
Let $u = 3^x$. Then $3^{2x} = (3^x)^2 = u^2$ [M1]:
$$u^2 - 10u + 9 = 0$$
Step 2: Factorise quadratic [1 mark]
$(u - 9)(u - 1) = 0 \implies u = 9 \text{ or } u = 1$ [A1]
Step 3: Solve for $x$ [2 marks]
• $3^x = 9 = 3^2 \implies x = 2$ [A1]
• $3^x = 1 = 3^0 \implies x = 0$ [A1]
Solutions: $x = 0, 2$.
In computer architecture, data storage units are defined in base 2:
• $1\text{ Kilobyte (KB)} = 2^{10}\text{ bytes}$
• $1\text{ Megabyte (MB)} = 2^{20}\text{ bytes}$
• $1\text{ Gigabyte (GB)} = 2^{30}\text{ bytes}$
(a) How many Kilobytes are in $1\text{ Gigabyte}$? Express your answer as a power of 2, then as an ordinary integer. [2 marks]
(b) A solid-state drive has capacity $64\text{ GB}$. Express its total byte capacity in standard form $a \times 10^k$ to 3 significant figures. [2 marks]
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(a) Kilobytes in 1 GB: [2 marks]
$\text{Ratio} = \frac{2^{30}\text{ bytes}}{2^{10}\text{ bytes}} = 2^{30 - 10} = \mathbf{2^{20}\text{ KB}}$ [A1]
$2^{20} = (2^{10})^2 = 1024^2 = \mathbf{1\,048\,576\text{ KB}}$ [A1]
(b) Total bytes of 64 GB SSD: [2 marks]
$64\text{ GB} = 2^6 \times 2^{30} = 2^{36}\text{ bytes}$ [M1]
$2^{36} = 68\,719\,476\,736\text{ bytes} \approx \mathbf{6.87 \times 10^{10}\text{ bytes}}$ [A1]
A strain of bacteria undergoes binary fission, doubling in population every $30\text{ minutes}$. An initial culture contains $N_0 = 500$ bacteria.
(a) Write a formula for the population $N(t)$ after $t$ hours. [2 marks]
(b) Calculate the population after $4\text{ hours}$. [1 mark]
(c) Determine after how many hours the culture population will reach $128\,000$. [2 marks]
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(a) Model formula: [2 marks]
In $t$ hours, the number of 30-minute intervals is $2t$.
$N(t) = 500 \times 2^{2t}$ [A2]
(b) Population at $t = 4$: [1 mark]
$N(4) = 500 \times 2^{2(4)} = 500 \times 2^8 = 500 \times 256 = \mathbf{128\,000\text{ bacteria}}$ [A1]
(c) Time to reach $128\,000$: [2 marks]
$500 \times 2^{2t} = 128\,000 \implies 2^{2t} = \frac{128\,000}{500} = 256 = 2^8$ [M1]
$2t = 8 \implies t = \mathbf{4\text{ hours}}$ [A1]
Index Laws Quick Reference
Operational Index Laws
Zero & Negative Powers
Distributive Laws
Base Equivalence Rule
Top 5 Common Student Traps
1. Confusing Powers of Negatives: $(-2)^4$ vs $-2^4$
$(-2)^4 = (-2)(-2)(-2)(-2) = +16$, but $-2^4 = -(2^4) = -16$. Brackets dictate whether the negative sign is part of the base!
2. Adding Bases in Product Rule
$2^3 \times 2^4 = 2^{3+4} = 2^7$. Students often write $4^7$ by multiplying the bases together. The base remains unchanged.
3. Misunderstanding $(a + b)^n$
$(a + b)^2 \ne a^2 + b^2$! Exponents distribute across multiplication and division, NOT across addition or subtraction.
4. Applying Negative Exponents to Coefficients
In $5x^{-2}$, only $x$ has the negative exponent: $5x^{-2} = \frac{5}{x^2}$. It is NOT $\frac{1}{5x^2}$ unless grouped as $(5x)^{-2}$.
Integer Exponents Check
Solve these 5 questions to test index laws, negative powers, and exponential equations.
Which expression is equivalent to $\frac{x^4 \times (x^2)^{-3}}{x^{-5}}$?
What is the exact value of $\left(\frac{2}{5}\right)^{-3}$?
If $3^{x+2} = \frac{1}{27}$, what is the value of $x$?
Solve for $x$: $4^{x-1} = 8^{2x+1}$.
Which expression simplifies $2^{n+1} + 2^n$ to a single term?