Power, Indices, and Exponents: These three terms are closely related and often used interchangeably in mathematics.
The exponent is the number that indicates how many times the base is multiplied by itself. In $a^n$, $a$ is called the base and $n$ is called the index or exponent .
For any non-zero real number $a$ and natural number $n$:
$$a^n = a \times a \times a \times \dots \times a \quad (n\text{ times})$$
$$a^1 = a, \quad a^2 = a \times a, \quad a^3 = a \times a \times a$$
$$\text{Example: } 2^4 = 2 \times 2 \times 2 \times 2 = 16$$
Special Powers & Boundary Conditions:
Zero Power: $a^0 = 1$ for any $a \ne 0$. (E.g. $4^0 = 1$). Note: $0^0$ is not defined and is called an indeterminate form. Negative Power: For positive $n$, $a^{-n} = \frac{1}{a^n}$ ($a \ne 0$). E.g. $2^{-1} = \frac{1}{2}$, $2^{-2} = \frac{1}{4}$, $2^{-3} = \frac{1}{8}$.Fractional Power: $a^{1/n} = \sqrt[n]{a}$. E.g. $9^{1/2} = \sqrt{9} = 3$, $8^{1/3} = \sqrt[3]{8} = 2$. More generally, $a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$.
Let $a$ and $b$ be non-zero real numbers, and let $m$ and $n$ be rational numbers. The operations of exponents satisfy the following governing laws:
1. Product Law:
$$a^m \cdot a^n = a^{m+n}$$
Example: $2^3 \cdot 2^2 = 2^{3+2} = 2^5 = 32$
2. Quotient Law:
$$\frac{a^m}{a^n} = a^{m-n}$$
Example: $\frac{2^3}{2^2} = 2^{3-2} = 2$, $\frac{2^2}{2^3} = 2^{2-3} = 2^{-1} = \frac{1}{2}$
3. Power of a Power Law:
$$(a^m)^n = a^{mn}$$
Example: $(2^2)^3 = 2^{2 \times 3} = 2^6 = 64$
4. Power of a Product Law:
$$(ab)^m = a^m b^m$$
Example: $3^2 \cdot 4^2 = (3 \cdot 4)^2 = 12^2 = 144$
5. Power of a Quotient Law:
$$\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} \quad (b \ne 0)$$
Example: $\frac{6^2}{2^2} = \left(\frac{6}{2}\right)^2 = 3^2 = 9$
Example 4 (a): Simplify $4(6^{\frac{2}{3}})^{\frac{3}{2}} + (4^3)^{\frac{4}{3} \times \dots} + 2 \times (3^5)^{\frac{1}{5}}$.
Show Step-by-Step Solution
Apply power of a power law $(a^m)^n = a^{mn}$ to each term:
Example 4 (b): Simplify $\sqrt{\frac{1}{4}} + (0.01)^{-1/2} - 27^{2/3}$.
Show Step-by-Step Solution
Evaluate each individual term:
Example 4 (c): Simplify $\frac{5^{n+2} - 6 \cdot 5^{n+1}}{13 \cdot 5^n - 2 \cdot 5^{n+1}}$.
Show Step-by-Step Solution
Factor out common base power $5^n$ from both numerator and denominator:
Example 4 (d): Simplify $16 - 3 \times 2^2 + \left(\left(\frac{6}{9}\right)^2\right)^{1/2}$.
Show Step-by-Step Solution
• Evaluate $3 \times 2^2 = 3 \times 4 = 12$
Next: Check Your Progress
Simplify the following expressions using the laws of exponents and indices. Click each solution toggle to verify your working.
Simplify the expression: $a = x^{\frac{1}{2}} \cdot x^{\frac{1}{3}}$.
Show Step-by-Step Solution
Step 1: Apply the Product Law: $a^m \cdot a^n = a^{m+n}$.Step 2: Find the common denominator for the exponents: $\frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6}$.Answer: $\mathbf{a = x^{5/6}}$.
Simplify the expression: $a = x^2 \cdot x^{\frac{2}{3}}$.
Show Step-by-Step Solution
Step 1: Apply Product Law: $a^m \cdot a^n = a^{m+n}$.Answer: $\mathbf{a = x^{8/3}}$.
Simplify: $a = \left(x^{\frac{2}{3}}\right)^6$.
Show Step-by-Step Solution
Step 1: Apply Power of a Power Law: $(a^m)^n = a^{mn}$.Answer: $\mathbf{a = x^4}$.
Simplify: $a = \frac{x^3 y^4}{x y^2}$.
Show Step-by-Step Solution
Step 1: Apply Quotient Law $\frac{a^m}{a^n} = a^{m-n}$ to each variable base independently:Answer: $\mathbf{a = x^2 y^2 = (xy)^2}$.
Next: Worksheet
Printable classroom and homework worksheet for Indices & Exponents . Selected questions directly from the CBSE Applied Mathematics support material.
Section A: Conceptual & Objective Drill
3 Questions
Evaluate the product: $6561^{0.14} \times 6561^{0.11}$.
Show Model Solution
By Product Law: $6561^{0.14 + 0.11} = 6561^{0.25} = 6561^{1/4}$.
Find the exact value of the difference: $7^{-14} - 7^{-15}$.
Show Model Solution
Factor out the smaller power $7^{-15}$:
Evaluate: $(6561)^{1/2} + (6561)^{1/4} + (6561)^{1/8}$.
Show Model Solution
• $(6561)^{1/2} = \sqrt{6561} = 81$
Section B: Short Numerical Problems
2 Questions • 2 Marks Each
If $3^{n+4} - 3^{n+2} = 8$, find the value of $n$.
Show Model Solution
Factor out $3^{n+2}$:
If $5^{a+b} = 5 \times 25 \times 125$, what is the value of $(a+b)^2$?
Show Model Solution
Write RHS in powers of 5: $5^1 \times 5^2 \times 5^3 = 5^{1+2+3} = 5^6$.
Section C: High-Yield Applied & Advanced Problems
2 Questions
If $36^{120} = (36 \times x)^{40}$, determine the value of $x$.
Show Model Solution
Rewrite LHS using power of a power: $36^{120} = (36^3)^{40}$.
Evaluate the exact radical expression: $\left[\frac{3^{-2} - 5^{-2}}{3^{-2} + 5^{-2}}\right]^{1/2}$.
Show Model Solution
Convert negative exponents: $3^{-2} = \frac{1}{9}$ and $5^{-2} = \frac{1}{25}$.
Next: Revision Sheet
Essential formulas, laws, and examination tips for rapid last-minute revision.
Law / Rule
Formula
Key Condition / Notes
Product Law $$a^m \cdot a^n = a^{m+n}$$ Same base, exponents add Quotient Law $$\frac{a^m}{a^n} = a^{m-n}$$ $a \ne 0$; subtract exponents Power of Power $$(a^m)^n = a^{mn}$$ Multiply exponents Product to Power $$(ab)^m = a^m b^m$$ Distributes across factors Quotient to Power $$\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$$ $b \ne 0$ Zero Exponent $$a^0 = 1$$ $a \ne 0$; $0^0$ is indeterminate Negative Exponent $$a^{-n} = \frac{1}{a^n}$$ $a \ne 0$ Rational / Fractional $$a^{m/n} = (\sqrt[n]{a})^m$$ $n$th root raised to power $m$
$(a+b)^n \ne a^n + b^n$. Exponents DO NOT distribute over addition or subtraction!
$a^{-n} \ne -a^n$. A negative exponent indicates a reciprocal, NOT a negative value (e.g. $2^{-3} = 1/8$, not $-8$).
Always reduce composite bases (like 6, 8, 27, 36) to prime factor powers ($2^p, 3^q$) before simplifying algebraic expressions!
Next: 10-Q Board Mock Test
Exercise 1.2 Diagnostic Board-Level Mock Examination
Exact pattern test covering Objective MCQs, Assertion-Reason, Short Answer Types, Long Answer, and a Case-Based Study.
Section A: Objective & Assertion-Reason (Q1 – Q4)
4 Questions • 1 Mark Each
What is the value of $6561^{0.14} \times 6561^{0.11}$?
Solution: By product law: $6561^{0.14 + 0.11} = 6561^{0.25} = 6561^{1/4} = (3^8)^{1/4} = 3^2 = 9$.
If $(\sqrt{3})^n = 6561$, then what is the value of $(n)^{1/2}$?
(A) 64
(B) $64\sqrt{3}$
(C) $16\sqrt{3}$
(D) 4
Solution: $(\sqrt{3})^n = 3^{n/2}$. Also $6561 = 3^8$. So $n/2 = 8 \implies n = 16$. Then $(n)^{1/2} = \sqrt{16} = 4$.
What is the value of $(7^{-14} - 7^{-15})$?
(A) $6 \times 7^{-15}$
(B) $6 \times 7^{-14}$
(C) $7 \times 7^{-15}$
(D) $7 \times 7^{-14}$
Solution: $7^{-14} - 7^{-15} = 7^{-15}(7^1 - 1) = 6 \times 7^{-15}$.
Assertion (A): For any non-zero real number $a$, $a^0 = 1$.Reason (R): By the quotient law of exponents, $\frac{a^m}{a^m} = a^{m-m} = a^0 = 1$.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Solution: Both A and R are true, and R provides the direct formal proof of why non-zero bases raised to power zero equal 1.
Section B: Short Answer Type I (Q5 – Q6)
2 Questions • 2 Marks Each
If $3^{n+4} - 3^{n+2} = 8$, find the value of $n$.
Show Marking Scheme & Solution (2 Marks)
Step 1 (1 Mark): Factor out $3^{n+2}$:Step 2 (1 Mark): Since $3^0 = 1$, equate exponents: $n + 2 = 0 \implies \mathbf{n = -2}$.
Simplify the expression: $\sqrt{\frac{1}{4}} + (0.01)^{-1/2} - 27^{2/3}$.
Show Marking Scheme & Solution (2 Marks)
Step 1 (1 Mark): Simplify each term:Step 2 (1 Mark): Combine: $1/2 + 10 - 9 = 1/2 + 1 = \mathbf{3/2}$.
Section C: Short Answer Type II (Q7 – Q8)
2 Questions • 3 Marks Each
Simplify $\frac{5^{n+2} - 6\cdot 5^{n+1}}{13\cdot 5^n - 2\cdot 5^{n+1}}$ showing complete factorisation steps.
Show Marking Scheme & Solution (3 Marks)
Step 1 (1 Mark): Factor $5^n$ from numerator and denominator:Step 2 (1 Mark): Cancel $5^n$ since $5^n \ne 0$: $\frac{25 - 30}{13 - 10}$.Step 3 (1 Mark): Simplify: $\frac{-5}{3} = -\mathbf{\frac{5}{3}}$.
If $36^{120} = (36 \times x)^{40}$, determine the numerical value of $x$ using exponent laws.
Show Marking Scheme & Solution (3 Marks)
Step 1 (1 Mark): Rewrite LHS: $36^{120} = (36^3)^{40}$.Step 2 (1 Mark): Equate bases since powers are equal: $36^3 = 36 \times x$.Step 3 (1 Mark): Solve for $x$: $x = \frac{36^3}{36} = 36^2 = \mathbf{1296}$.
Section D: Long Answer Type (Q9)
1 Question • 4 Marks
Evaluate $\left[\frac{3^{-2} - 5^{-2}}{3^{-2} + 5^{-2}}\right]^{1/2}$ and simplify $\frac{2^{n+1} + 2^n}{2^{n+2} - 2^{n+1}}$.
Show Marking Scheme & Solution (4 Marks)
Part 1 [2 Marks]: $3^{-2} = 1/9, 5^{-2} = 1/25$. Numerator $= 1/9 - 1/25 = 16/225$. Denominator $= 1/9 + 1/25 = 34/225$. Ratio $= 16/34 = 8/17$. Square root $= \sqrt{8/17} = \mathbf{\frac{2\sqrt{2}}{\sqrt{17}}}$.Part 2 [2 Marks]: Factor $2^n$: $\frac{2^n(2 + 1)}{2^n(4 - 2)} = \frac{3}{2} = \mathbf{1.5}$.
Section E: Case-Based / Source-Based Problem (Q10)
1 Case Study • 4 Marks
Case Background: In a biological research facility, a bacterial culture follows an exponential growth model given by $N(t) = 500 \cdot 2^{t/3}$, where $t$ is measured in hours.
(i) [1 Mark]: Find the initial bacteria population at $t = 0$.
Show Solution for (i)
[1 Mark]: At $t = 0$: $N(0) = 500 \cdot 2^0 = 500 \times 1 = \mathbf{500}$ bacteria.
(ii) [1 Mark]: Determine the bacteria population after 6 hours.
Show Solution for (ii)
[1 Mark]: At $t = 6$: $N(6) = 500 \cdot 2^{6/3} = 500 \cdot 2^2 = 500 \times 4 = \mathbf{2000}$ bacteria.
(iii) [2 Marks]: Calculate the time $t$ in hours when the bacteria count will reach 16,000.
Show Solution for (iii)
[2 Marks]: Set $N(t) = 16000$: $500 \cdot 2^{t/3} = 16000 \implies 2^{t/3} = 32$. Since $32 = 2^5$: $t/3 = 5 \implies \mathbf{t = 15\text{ hours}}$.