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ICSE • Class 9 • Mathematics • Ch 6
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Indices

In ICSE Class 9 Mathematics, "Indices" (also universally designated as Exponents or Powers) provides the formal algebraic language for expressing repeated multiplication, fractional roots, scientific notation, and exponential growth/decay. For any base $a \in \mathbb{R}$ and exponent $n \in \mathbb{N}$, the expression $a^n$ denotes the product of $n$ factors of $a$. The chapter rigorously extends this definition from positive natural numbers to the entire rational number set $\mathbb{Q}$, covering zero exponents ($a^0 = 1$ for $a \neq 0$), negative integral exponents ($a^{-n} = \frac{1}{a^n}$), and fractional exponents ($a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$). Students master the fundamental Laws of Indices: the Product Law ($a^m \times a^n = a^{m+n}$), Quotient Law ($a^m \div a^n = a^{m-n}$), Power of a Power Law ($(a^m)^n = a^{mn}$), Power of a Product Law ($(ab)^n = a^n b^n$), and Power of a Quotient Law (($a/b)^n = a^n / b^n$). Crucial analytical applications include prime factor decomposition of bases, simplifying nested fractional radical fractions without evaluating massive numbers directly, and solving exponential equations where variables appear in the exponents by invoking the fundamental Bi-conditional Base Identity: if $a^x = a^y$ and $a \notin \{-1, 0, 1\}$, then $x = y$.

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The Legend of the Chessboard: How 64 Squares Bankrupted an Ancient King with Powers of Two

In ancient India, King Shirham was so delighted by the newly invented game of chess that he summoned its creator, the sage Sissa, and offered him any royal reward he desired. Sissa humbly made a seemingly modest request: "Your Majesty, place a single grain of rice on the first square of the board, two grains on the second, four on the third, eight on the fourth, and keep doubling the grains for each of the 64 squares." The King laughed at what he thought was a foolishly small request and ordered his treasurer to fetch a sack of rice. But by the 30th square, the royal storehouses were emptied! By the 64th square, the required number was $2^{63}$ grains, bringing the total sum to $2^{64} - 1 = 18,446,744,073,709,551,615$ grains of rice—over 500 billion metric tons, enough to blanket the entire surface of planet Earth in a meter-deep layer of rice! That is the mind-bending power of Indices. How do exponents allow modern scientists to write the mass of an electron ($9.11 \times 10^{-31}\text{ kg}$) or calculate the processing speed of a supercomputer? Let us master the laws of exponents!

Why This Chapter Matters

Indices are universal in physics (Newtonian gravitation, atomic physics, radioactive half-life), computer science (binary algorithms, bit allocation), and finance (continuous compounding).

Before You Begin (Prerequisites)

  • Prime factorisation of natural numbers.
  • Basic arithmetic of fractions and negative integers.

What You Will Learn (Core Objectives)

  • State and prove the universal laws of indices for positive, zero, negative, and fractional powers.
  • Simplify complex rational expressions involving mixed exponential bases by prime factorisation.
  • Convert between radical form $\sqrt[n]{a^m}$ and exponential form $a^{m/n}$.
  • Solve exponential equations by equating exponents when bases are identical.
  • Solve simultaneous exponential equations involving multiple unknown powers.

Chapter Roadmap & Progression

1 1. Fundamental Laws of Indices
2 2. Exponential Equations & Equating...
3 3. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamental Laws of Indices

Universal Exponent Laws

For any non-zero real bases $a, b$ and rational exponents $m, n$:

Law NameMathematical FormulaCondition / Meaning
Product Law$a^m \times a^n = a^{m+n}$Same base multiplied $\implies$ add exponents.
Quotient Law$\frac{a^m}{a^n} = a^{m-n}$Same base divided $\implies$ subtract exponents.
Power of a Power$(a^m)^n = a^{mn} = (a^n)^m$Powers stacked $\implies$ multiply exponents.
Power of a Product$(ab)^n = a^n \times b^n$Exponent distributes over multiplication.
Power of a Quotient$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$Exponent distributes over division ($b \neq 0$).
Zero Exponent$a^0 = 1$Any non-zero base to power zero is $1$ ($a \neq 0$).
Negative Exponent$a^{-n} = \frac{1}{a^n} \iff \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$Negative power inverts the base fraction.
Fractional Exponent$a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$Denominator $n$ is root index, numerator $m$ is power.

2. Exponential Equations & Equating Powers Principle

Solving Equations
The Equating Powers Theorem:
$$\mathbf{\text{If } a^x = a^y \quad \text{and } a \notin \{-1, 0, 1\}, \quad \text{then } x = y}$$

To solve an exponential equation where the variable is locked in the exponent:

  1. Express all composite numbers on both sides in terms of their prime base factors (e.g., replace $4$ with $2^2$, $9$ with $3^2$, $27$ with $3^3$, $32$ with $2^5$).
  2. Apply the laws of indices to combine terms into a single base on each side: $a^{f(x)} = a^{g(x)}$.
  3. Equate the exponents: $f(x) = g(x)$ and solve the resulting algebraic equation.

3. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Simplify:
$$\frac{9^n \times 3^2 \times (3^{-n/2})^{-2} - (27)^n}{3^{3m} \times 2^3} = \frac{1}{27}, \quad \text{prove that } m - n = 1.$$

Solution:

Express all terms in base $3$:

  • $9^n = (3^2)^n = 3^{2n}$
  • $(3^{-n/2})^{-2} = 3^{(-n/2) \times (-2)} = 3^n$
  • $27^n = (3^3)^n = 3^{3n}$
  • $27 = 3^3 \implies \frac{1}{27} = 3^{-3}$

Substitute into the numerator:

$$\text{Numerator} = 3^{2n} \times 3^2 \times 3^n - 3^{3n} = 3^{2n + 2 + n} - 3^{3n} = 3^{3n+2} - 3^{3n}$$ $$= 3^{3n}(3^2 - 1) = 3^{3n}(9 - 1) = 3^{3n} \times 8$$

Now simplify the complete fraction:

$$\frac{3^{3n} \times 8}{3^{3m} \times 8} = \frac{3^{3n}}{3^{3m}} = 3^{3n - 3m}$$

Equate this to the right-hand side:

$$3^{3n - 3m} = \frac{1}{27} = 3^{-3}$$

Equating exponents since the base ($3$) is identical:

$$3n - 3m = -3 \implies 3(m - n) = 3 \implies \mathbf{m - n = 1} \quad \blacksquare$$
Problem 2: Solve for $x$: $2^{2x+3} - 9 \cdot 2^x + 1 = 0$.

Solution:

Split the first term using $2^{2x+3} = 2^{2x} \cdot 2^3 = 8 \cdot (2^x)^2$:

$$8(2^x)^2 - 9(2^x) + 1 = 0$$

Let $y = 2^x$. The equation becomes a quadratic in $y$:

$$8y^2 - 9y + 1 = 0$$

Factorise by splitting the middle term ($ac = 8$, sum = $-9 \implies -8$ and $-1$):

$$8y^2 - 8y - y + 1 = 0 \implies 8y(y - 1) - 1(y - 1) = 0$$ $$(y - 1)(8y - 1) = 0 \implies y = 1 \quad \text{or} \quad y = \frac{1}{8}$$

Substitute back $y = 2^x$:

  • Case 1: $2^x = 1 = 2^0 \implies \mathbf{x = 0}$
  • Case 2: $2^x = \frac{1}{8} = \frac{1}{2^3} = 2^{-3} \implies \mathbf{x = -3}$
$$\mathbf{\text{Solutions: } x = 0, \quad x = -3}$$

Key Formulas, Identities & Theorems

Product Law
$$a^m \times a^n = a^{m+n}$$
Add exponents when multiplying identical bases.
Quotient Law
$$\frac{a^m}{a^n} = a^{m-n}$$
Subtract exponents when dividing identical bases.
Power of Power
$$(a^m)^n = a^{mn}$$
Multiply exponents when raised to another power.
Zero Exponent
$$a^0 = 1, \quad a \neq 0$$
Any non-zero real number to power 0 equals 1.
Negative Power Inversion
$$a^{-n} = \frac{1}{a^n}, \quad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$$
Inverts the base fraction.
Fractional Exponent
$$a^{m/n} = \sqrt[n]{a^m}$$
Root index n, power m.

Mathematics: Laws of Indices & Exponential Curve Growth

Laws of Indices & Exponential Curves: y = 2ˣ vs y = (1/2)ˣ Exponential Growth vs. Decay X Y (0, 1) y = 2ˣ y = 2⁻ˣ Horizontal Asymptote: y = 0 (X-axis) Summary of Index Rules Multiplication & Division: aᵐ × aⁿ = aᵐ⁺ⁿ  |  aᵐ ÷ aⁿ = aᵐ⁻ⁿ Powers & Inverses: (aᵐ)ⁿ = aᵐⁿ  |  a⁻ⁿ = 1 / aⁿ Zero & Fractions: a⁰ = 1 (a ≠ 0)  |  aᵐ/ⁿ = ⁿ√(aᵐ) Solving Exponential Equations: 1. Reduce all bases to primes (2, 3, 5...) 2. Combine using product/quotient laws 3. If aˣ = aʸ (a ≠ 0, ±1) ⇒ x = y

Chapter Summary & 10 Key Takeaways

Takeaway 1
An index (exponent) denotes repeated multiplication of a base: a^n = a * a * ... * a (n times).
Takeaway 2
Product rule: a^m * a^n = a^(m+n).
Takeaway 3
Quotient rule: a^m / a^n = a^(m-n).
Takeaway 4
Power of power rule: (a^m)^n = a^(mn).
Takeaway 5
Any non-zero real number raised to power 0 equals 1: a^0 = 1.
Takeaway 6
Negative exponent rule: a^(-n) = 1/a^n and (a/b)^(-n) = (b/a)^n.
Takeaway 7
Fractional exponent rule: a^(m/n) = nth root of a^m.
Takeaway 8
To simplify complex exponential expressions, decompose all composite bases into their prime factors.
Takeaway 9
Exponential equations of the form a^f(x) = a^g(x) are solved by setting f(x) = g(x) when a ≠ 0, 1, -1.
Takeaway 10
Equations with terms like 2^(2x) and 2^x can be converted to quadratic equations by setting y = 2^x.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Simplify: $\frac{2^{n} + 2^{n-1}}{2^{n+1} - 2^n}$.
Reveal Answer & Explanation
Answer: • Factor out $2^n$ from both the numerator and the denominator:
$$\text{Numerator} = 2^n + 2^n \cdot 2^{-1} = 2^n\left(1 + \frac{1}{2}\right) = 2^n\left(\frac{3}{2}\right)$$
$$\text{Denominator} = 2^n \cdot 2^1 - 2^n = 2^n(2 - 1) = 2^n(1)$$
• Dividing numerator by denominator:
$$\frac{2^n\left(\frac{3}{2}\right)}{2^n(1)} = \frac{3}{2} = \mathbf{1.5}$$
Factor out 2^n from both numerator and denominator.
2
Solve for $x$: $4^{x-1} \times (0.5)^{3-2x} = \left(\frac{1}{8}\right)^x$.
Reveal Answer & Explanation
Answer: • Express all bases as powers of $2$:
- $4 = 2^2 \implies 4^{x-1} = (2^2)^{x-1} = 2^{2x-2}$
- $0.5 = \frac{1}{2} = 2^{-1} \implies (0.5)^{3-2x} = (2^{-1})^{3-2x} = 2^{-(3-2x)} = 2^{2x-3}$
- $\frac{1}{8} = \frac{1}{2^3} = 2^{-3} \implies \left(\frac{1}{8}\right)^x = 2^{-3x}$
• Combine the LHS using the product law:
$$\text{LHS} = 2^{2x-2} \times 2^{2x-3} = 2^{(2x-2) + (2x-3)} = 2^{4x - 5}$$
• Equate LHS to RHS:
$$2^{4x - 5} = 2^{-3x}$$
• Equating exponents since the bases are identical ($2$):
$$4x - 5 = -3x \implies 7x = 5 \implies \mathbf{x = \frac{5}{7}}$$
Convert 4 to 2^2, 0.5 to 2^(-1), and 1/8 to 2^(-3). Then equate powers of 2.
3
If $2^x = 3^y = 6^{-z}$, prove that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.
Reveal Answer & Explanation
Answer: • Let $2^x = 3^y = 6^{-z} = k$, where $k$ is a non-zero constant.
• Express each base in terms of $k$:
$$2 = k^{1/x}, \quad 3 = k^{1/y}, \quad 6 = k^{-1/z}$$
• We know the arithmetic relation between the bases: $2 \times 3 = 6$.
• Substitute the powers of $k$:
$$k^{1/x} \times k^{1/y} = k^{-1/z}$$
$$k^{\frac{1}{x} + \frac{1}{y}} = k^{-\frac{1}{z}}$$
• Equating the exponents:
$$\frac{1}{x} + \frac{1}{y} = -\frac{1}{z} \implies \mathbf{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0} \quad \blacksquare$$
Set 2^x = 3^y = 6^(-z) = k. Use 2 * 3 = 6.
4
Evaluate: $\left(\frac{81}{16}\right)^{-3/4} \times \left[\left(\frac{25}{9}\right)^{-3/2} \div \left(\frac{5}{2}\right)^{-3}\right]$.
Reveal Answer & Explanation
Answer:

• Step 1: Simplify $\left(\frac{81}{16}\right)^{-3/4}$:

$$\frac{81}{16} = \left(\frac{3}{2}\right)^4 \implies \left[\left(\frac{3}{2}\right)^4\right]^{-3/4} = \left(\frac{3}{2}\right)^{-3} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$$


• Step 2: Simplify $\left(\frac{25}{9}\right)^{-3/2}$:

$$\frac{25}{9} = \left(\frac{5}{3}\right)^2 \implies \left[\left(\frac{5}{3}\right)^2\right]^{-3/2} = \left(\frac{5}{3}\right)^{-3} = \left(\frac{3}{5}\right)^3 = \frac{27}{125}$$


• Step 3: Simplify $\left(\frac{5}{2}\right)^{-3}$:

$$\left(\frac{5}{2}\right)^{-3} = \left(\frac{2}{5}\right)^3 = \frac{8}{125}$$


• Step 4: Evaluate the bracket: $\frac{27}{125} \div \frac{8}{125} = \frac{27}{125} \times \frac{125}{8} = \frac{27}{8}$
• Step 5: Multiply by the first term:

$$\frac{8}{27} \times \frac{27}{8} = \mathbf{1}$$


Rewrite 81/16 as (3/2)^4 and 25/9 as (5/3)^2. The product simplifies to 1.
5
Solve for $x$: $3^{x+1} + 3^{2-x} = 28$.
Reveal Answer & Explanation
Answer:

• Rewrite using index laws: $3 \cdot 3^x + \frac{3^2}{3^x} = 28 \implies 3 \cdot 3^x + \frac{9}{3^x} = 28$.
• Let $y = 3^x$ (where $y > 0$):

$$3y + \frac{9}{y} = 28 \implies 3y^2 - 28y + 9 = 0$$


• Split the middle term ($ac = 27$, sum = $-28 \implies -27$ and $-1$):

$$3y(y - 9) - 1(y - 9) = 0 \implies (y - 9)(3y - 1) = 0$$


$$y = 9 \quad \text{or} \quad y = \frac{1}{3}$$


• Substitute back $y = 3^x$:
- $3^x = 9 = 3^2 \implies \mathbf{x = 2}$
- $3^x = \frac{1}{3} = 3^{-1} \implies \mathbf{x = -1}$
• Solutions: $\mathbf{x = 2, \quad x = -1}$.


Substitute y = 3^x to get 3y + 9/y = 28. Solve 3y^2 - 28y + 9 = 0.
6
Show that: $\frac{1}{1 + x^{a-b}} + \frac{1}{1 + x^{b-a}} = 1$.
Reveal Answer & Explanation
Answer: • Using the negative index rule $x^{b-a} = \frac{1}{x^{a-b}}$:
$$\frac{1}{1 + x^{b-a}} = \frac{1}{1 + \frac{1}{x^{a-b}}} = \frac{1}{\frac{x^{a-b} + 1}{x^{a-b}}} = \frac{x^{a-b}}{x^{a-b} + 1}$$
• Now add both fractions:
$$\text{LHS} = \frac{1}{1 + x^{a-b}} + \frac{x^{a-b}}{1 + x^{a-b}} = \frac{1 + x^{a-b}}{1 + x^{a-b}} = \mathbf{1} = \text{RHS} \quad \blacksquare$$
Write x^(b-a) as 1 / x^(a-b) and simplify the second fraction.
7
If $a = x^{p-1}, b = x^{q-1}, c = x^{r-1}$, evaluate $a^{q-r} \cdot b^{r-p} \cdot c^{p-q}$.
Reveal Answer & Explanation
Answer: • Substitute the values of $a, b, c$ into the expression:
$$a^{q-r} = (x^{p-1})^{q-r} = x^{(p-1)(q-r)}$$
$$b^{r-p} = (x^{q-1})^{r-p} = x^{(q-1)(r-p)}$$
$$c^{p-q} = (x^{r-1})^{p-q} = x^{(r-1)(p-q)}$$
• Multiplying the terms using the product law gives $x$ raised to the sum of the exponents:
$$\text{Sum} = (pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q)$$
$$= (pq - pq) + (-pr + pr) + (-q + q) + (r - r) + (qr - qr) + (p - p) = 0$$
• Therefore:
$$\text{Value} = x^0 = \mathbf{1}$$
Apply power of power rule and sum exponents. All terms cancel cyclically to 0, giving x^0 = 1.
8
Solve the simultaneous exponential equations: $2^{x+y} = 32$ and $3^{3x-y} = 27$.
Reveal Answer & Explanation
Answer:

• Express RHS in prime powers:
$32 = 2^5 \implies 2^{x+y} = 2^5 \implies x + y = 5$   --- (1)
$27 = 3^3 \implies 3^{3x-y} = 3^3 \implies 3x - y = 3$   --- (2)
• Add Equation (1) and Equation (2):

$$(x + y) + (3x - y) = 5 + 3 \implies 4x = 8 \implies \mathbf{x = 2}$$


• Substitute $x = 2$ into (1):

$$2 + y = 5 \implies \mathbf{y = 3}$$


• Solution: $\mathbf{x = 2, \quad y = 3}$.


Set x + y = 5 and 3x - y = 3. Add to find x = 2.
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