Official Syllabus Item: Part of 46
Algebra & Equations
Exponents & Powers (Laws of Indices)
Master laws of exponents, negative indices, prime factorization of composite bases, exponential equations, and scientific notation for RIMC Entrance Exam.
Overview & Fundamental Concept
Exponents and Powers (Laws of Indices) form an essential component of the Algebra section of the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 46 of Chapter 8 in the official RIMC brochure, questions on exponents evaluate a candidate's mastery over index laws, manipulation of negative and fractional powers, decomposition of composite bases into prime factor products ($2, 3, 5, 7$), solving algebraic equations with unknown indices, and formatting extreme physical values into standard scientific notation ($m \times 10^k$). Descriptive step marking strictly requires writing the formal exponential law in parentheses at each reduction stage, showing prime factor trees, avoiding negative powers in final results, and boxing answers with absolute precision.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Part of Item 46 | Exponents and indices (simple problems), scientific notation, and index laws | Official brochure syllabus item. Tests foundational algebraic manipulation, base harmonization, exponential equation solving, and dimensional representation. |
1. Foundational Anatomy of Powers & Exponential Notation
Repeated addition of the same number leads to multiplication: $a + a + a + a = 4a$. Similarly, repeated multiplication of a number by itself leads to exponentiation: $a \times a \times a \times a = a^4$.
In the expression $a^n$:
• $a$ is the Base: The number being repeatedly multiplied.
• $n$ is the Exponent (or Index / Power): The number of times the base is used as a factor.
• $a^n$ is read as: '$a$ raised to the power $n$' or 'the $n$-th power of $a$'.
When the base is negative, parity of the exponent dictates the sign:
$$(-a)^n = \begin{cases} +a^n, & \text{if } n \text{ is an even integer} \\ -a^n, & \text{if } n \text{ is an odd integer} \end{cases}$$
Example: $(-3)^4 = +81$, whereas $(-3)^3 = -27$. Note the critical distinction: $-3^4 = -(3^4) = -81$, whereas $(-3)^4 = +81$.
Zero and Negative Exponents
- Zero Exponent Law: For any non-zero real number $a \neq 0$, $a^0 = 1$. This arises directly from the quotient law: $\frac{a^n}{a^n} = a^{n-n} = a^0 = 1$. Note: $0^0$ is indeterminate and undefined in basic arithmetic.
- Negative Exponent Law: A negative exponent indicates the reciprocal of the positive power: $a^{-n} = \frac{1}{a^n}$ and $\frac{1}{a^{-n}} = a^n$.
- Reciprocal of a Fraction with Negative Exponent: $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ (where $a, b \neq 0$).
2. The Fundamental Laws of Indices (Integral Exponents)
In RIMC descriptive papers, every simplification step must cite the relevant law in the margin. The table below outlines the core laws applicable for non-zero bases $a, b$ and integers $m, n$:
| Law Name | Mathematical Identity | Descriptive Exam Application |
|---|---|---|
| Product Law | $a^m \times a^n = a^{m+n}$ | When bases are identical, add powers: $2^3 \times 2^5 = 2^{3+5} = 2^8$. |
| Quotient Law | $a^m \div a^n = a^{m-n} = \frac{a^m}{a^n}$ | When bases are identical, subtract denominator power from numerator power. |
| Power of a Power | $(a^m)^n = a^{m \times n} = a^{mn}$ | Multiply inner and outer exponents: $(3^2)^4 = 3^8$. |
| Power of a Product | $(a \times b)^n = a^n \times b^n$ | Distribute exponent across factors: $(2 \times 5)^3 = 2^3 \times 5^3$. |
| Power of a Quotient | $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ | Distribute exponent across numerator and denominator. |
| Negative Power Rule | $a^{-n} = \frac{1}{a^n},\; \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ | Invert the base to make the power positive. |
| Zero Power Identity | $a^0 = 1 \quad (a \neq 0)$ | Any non-zero quantity raised to power $0$ equals $1$. |
3. Strategy for Simplifying Complex Exponential Expressions
Examiners frequently design questions containing composite numbers like $4, 6, 8, 9, 10, 12, 15, 25, 27$. The golden rule for solving such questions in RIMC is Universal Prime Base Harmonization:
The 4-Step Prime Harmonization Protocol
- Step 1 (Factorize to Primes): Express every composite base as a product of prime powers: $4 = 2^2$, $6 = 2 \times 3$, $8 = 2^3$, $9 = 3^2$, $10 = 2 \times 5$, $12 = 2^2 \times 3$, $25 = 5^2$, $27 = 3^3$.
- Step 2 (Distribute Exponents): Apply $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$ to eliminate all grouping parentheses.
- Step 3 (Collect Like Bases): Combine all powers of $2$, powers of $3$, powers of $5$, etc., in the numerator and denominator separately using $a^m \times a^n = a^{m+n}$.
- Step 4 (Quotient Reduction): Apply $\frac{a^m}{a^n} = a^{m-n}$ for each prime base to arrive at the irreducible integer or fraction.
Cadet Prime Base Lookup Table
Commit to memory: $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128, 2^8=256, 2^9=512, 2^{10}=1024$. For 3: $3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729$. For 5: $5^1=5, 5^2=25, 5^3=125, 5^4=625$.
4. Solving Exponential Equations (Unknown Index)
In RIMC Part B (10-mark problems), candidates are tasked with finding the value of an unknown variable $x$ or $m$ located in the exponent.
The governing mathematical theorem is the One-to-One Property of Exponential Functions:
$$\text{If } a^x = a^y \quad (\text{where } a > 0,\; a \neq 1), \quad \text{then } x = y$$
If the bases on both sides of the equation are equal, their exponents must be strictly equal. Conversely, if $x^n = y^n$ (with $n$ odd and $n \neq 0$), then $x = y$.
5. Standard Scientific Notation (Standard Form of Large & Small Numbers)
A number is written in Scientific Notation (Standard Form) when expressed in the form:
$$m \times 10^k$$
where $m$ is a decimal number such that $1.0 \le m < 10$ (exactly one non-zero digit before the decimal point) and $k$ is an integer $(\mathbb{Z})$.
Conversion Rules for Standard Form
- Very Large Numbers $(k > 0)$: Shift the decimal point to the left until there is only one non-zero digit to its left. The positive power $k$ equals the number of places shifted.
Example: $149,600,000,000\text{ m} = 1.496 \times 10^{11}\text{ m}$ (shifted 11 places left). - Very Small Decimals $(k < 0)$: Shift the decimal point to the right until it is immediately after the first non-zero digit. The negative power $-k$ equals the number of places shifted.
Example: $0.000007\text{ m} = 7 \times 10^{-6}\text{ m}$ (shifted 6 places right).
Illustrated Conceptual Diagrams & Visual Schematics
Comprehensive Laws of Exponents & Index Architecture
The 6 Universal Laws of Indices: (1) Product Law: $a^m \cdot a^n = a^{m+n}$. (2) Quotient Law: $a^m / a^n = a^{m-n}$. (3) Power of a Power: $(a^m)^n = a^{mn}$. (4) Power of a Product: $(ab)^n = a^n b^n$. (5) Zero Power: $a^0 = 1$ ($a \neq 0$). (6) Negative Exponent: $a^{-n} = \frac{1}{a^n}$.
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: Base Factorization & Prime Decomposition
1 - 2 Marks
Convert all composite numeric bases into prime powers (e.g., $10 \to 2 \times 5$, $25 \to 5^2$).
Booklet Layout:
Given: $\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$\nExpressing composite bases as prime powers:\n$25 = 5^2$ and $10 = 2 \times 5$.
Given: $\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$\nExpressing composite bases as prime powers:\n$25 = 5^2$ and $10 = 2 \times 5$.
Step 2: Citation of Governing Index Laws
1 - 2 Marks
Explicitly quote the identity used (e.g., $a^m \times a^n = a^{m+n}$, $(ab)^n = a^n b^n$).
Booklet Layout:
Using the Product Law: $a^m \times a^n = a^{m+n}$\nand Quotient Law: $\frac{a^m}{a^n} = a^{m-n}$
Using the Product Law: $a^m \times a^n = a^{m+n}$\nand Quotient Law: $\frac{a^m}{a^n} = a^{m-n}$
Step 3: Algebraic Power Arithmetic & Sign Reduction
2 - 4 Marks
Show arithmetic addition and subtraction of powers per prime base carefully handling negative signs.
Booklet Layout:
$$= 5^{2 - (-3) - 1} \times 2^{-1} \times t^{-4 - (-8)} = 5^{2 + 3 - 1} \times \frac{1}{2} \times t^{-4 + 8} = 5^4 \times \frac{1}{2} \times t^4$$
$$= 5^{2 - (-3) - 1} \times 2^{-1} \times t^{-4 - (-8)} = 5^{2 + 3 - 1} \times \frac{1}{2} \times t^{-4 + 8} = 5^4 \times \frac{1}{2} \times t^4$$
Step 4: Irreducible Final Value & Boxed Concluding Statement
1 Mark
Calculate numerical values ($5^4 = 625$) and clearly box the answer with units or conditions.
Booklet Layout:
$$= \frac{625 t^4}{2} \quad (t \neq 0)$$
$$\mathbf{\text{Answer: } \frac{625}{2}t^4}$$
$$= \frac{625 t^4}{2} \quad (t \neq 0)$$
$$\mathbf{\text{Answer: } \frac{625}{2}t^4}$$
Examiner Traps & Common Blunders (Marks Penalties)
Multiplying Bases when Multiplying Powers -2 Marks
Common Mistake: Writing $2^3 \times 2^4 = 4^7$ or $2^3 \times 2^4 = 4^{12}$.
Cadet Antidote: The base never changes under the Product Law: $a^m \times a^n = a^{m+n}$. Thus, $2^3 \times 2^4 = 2^{3+4} = 2^7 = 128$.
Confusing Exponentiation with Multiplication -2 Marks
Common Mistake: Computing $3^4$ as $3 \times 4 = 12$.
Cadet Antidote: $3^4$ means $3 \times 3 \times 3 \times 3 = 81$. Exponents indicate repeated multiplication, not multiplication by the power.
Misinterpreting Parentheses with Negative Bases -1 to -2 Marks
Common Mistake: Treating $-5^2$ as $+25$.
Cadet Antidote: $-5^2 = -(5 \times 5) = -25$. The power applies only to the base 5. If the negative sign is included, brackets are mandatory: $(-5)^2 = (-5) \times (-5) = +25$.
Scientific Notation Non-Compliant Coefficient -1 Mark
Common Mistake: Writing $35.4 \times 10^5$ or $0.354 \times 10^7$ as standard scientific form.
Cadet Antidote: The coefficient $m$ must satisfy $1.0 \le m < 10$. Exactly one non-zero digit must precede the decimal point: $3.54 \times 10^6$.
Cadet Topper Mental Verification Shortcuts
Instant Base-Equating for Exponential Equations
Whenever $a^{f(x)} = b$, immediately express $b$ as a power of $a$. If bases differ, factor both sides into prime bases and equate corresponding exponents independently.
If $2^{x-1} + 2^{x+1} = 320$, factor out $2^x$: $2^x(2^{-1} + 2^1) = 2^x(\frac{1}{2} + 2) = 2^x(\frac{5}{2}) = 320 \implies 2^x = 320 \times \frac{2}{5} = 128 = 2^7 \implies x = 7$.
Quick Decimal Shift Count for Scientific Notation
For small decimals: count all zeros after the decimal point up to the first non-zero digit, then add 1. That total gives the negative power of 10.
In $0.000045$: there are 4 zeros between decimal and 4. $4 + 1 = 5$, so $4.5 \times 10^{-5}$ instantly.
Fraction Inversion for Negative Indices
Flip the fraction instantly to change the sign of the exponent from negative to positive before doing any computation.
$\left(\frac{2}{5}\right)^{-3} = \left(\frac{5}{2}\right)^3 = \frac{125}{8}$.
Key Formulas & Conversion Cheat Sheet
Product of Powers with Same Base
$$a^m \times a^n = a^{m+n}$$
Bases must be identical; exponents are added.
Quotient of Powers with Same Base
$$a^m \div a^n = \frac{a^m}{a^n} = a^{m-n} \quad (a \neq 0)$$
Bases must be identical; denominator power is subtracted from numerator power.
Power of a Power
$$(a^m)^n = a^{m \times n} = a^{mn}$$
Powers in cascade are multiplied.
Power of a Product
$$(a \times b)^n = a^n \times b^n$$
The exponent distributes over each multiplying factor.
Power of a Quotient
$$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (b \neq 0)$$
The exponent distributes over numerator and denominator.
Negative Exponent Reciprocal Rule
$$a^{-n} = \frac{1}{a^n} \quad \text{and} \quad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$$
Inverting the fraction changes the sign of the exponent.
Zero Exponent Identity
$$a^0 = 1 \quad (a \neq 0)$$
Any non-zero real number raised to the zero power equals 1.
Standard Scientific Notation
$$N = m \times 10^k \quad (1.0 \le m < 10,\; k \in \mathbb{Z})$$
Standard representation for astronomical or microscopic measurements.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Nested Negative Exponents
5 Marks
Evaluate the following expression step-by-step:
$$\left[ \left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1} \right]^{-1}$$
$$\left[ \left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1} \right]^{-1}$$
Step-by-Step Model Solution:
Step 1: Apply Negative Exponent Rule to Inner Fractions
2 Marks
We use the rule $\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}$.
$$\left(\frac{1}{3}\right)^{-1} = \frac{3}{1} = 3$$
$$\left(\frac{1}{4}\right)^{-1} = \frac{4}{1} = 4$$
$$\left(\frac{1}{3}\right)^{-1} = \frac{3}{1} = 3$$
$$\left(\frac{1}{4}\right)^{-1} = \frac{4}{1} = 4$$
Rough Column Working: (1/a)^(-1) = a
Step 2: Evaluate Inside the Square Brackets
1 Mark
Substitute the evaluated inner values:
$$\left[ 3 - 4 \right]^{-1} = \left[ -1 \right]^{-1}$$
$$\left[ 3 - 4 \right]^{-1} = \left[ -1 \right]^{-1}$$
Step 3: Evaluate Outer Negative Exponent
2 Marks
Using the reciprocal rule $a^{-1} = \frac{1}{a}$:
$$\left[ -1 \right]^{-1} = \frac{1}{-1} = -1$$
$$\mathbf{\text{Final Answer: } -1}$$
$$\left[ -1 \right]^{-1} = \frac{1}{-1} = -1$$
$$\mathbf{\text{Final Answer: } -1}$$
Examiner Scoring Key: Award 2 marks for inverting inner fractions, 1 mark for obtaining -1 inside brackets, and 2 marks for evaluating (-1)^(-1) = -1.
Model 2: Solving Exponential Equation
5 Marks
Find the value of $m$ for which:
$$5^m \div 5^{-3} = 5^5$$
$$5^m \div 5^{-3} = 5^5$$
Step-by-Step Model Solution:
Step 1: Apply Quotient Law of Exponents
2 Marks
The given equation is $5^m \div 5^{-3} = 5^5$.
Using the Quotient Law: $a^m \div a^n = a^{m - n}$ (where base $a = 5$):
$$5^{m - (-3)} = 5^5$$
$$5^{m + 3} = 5^5$$
Using the Quotient Law: $a^m \div a^n = a^{m - n}$ (where base $a = 5$):
$$5^{m - (-3)} = 5^5$$
$$5^{m + 3} = 5^5$$
Step 2: Equate Exponents of Equal Bases
2 Marks
Since the bases on both sides of the equation are equal ($5 = 5$) and $5 \neq 0, 1, -1$, their exponents must be equal:
$$m + 3 = 5$$
$$m + 3 = 5$$
Step 3: Solve Simple Linear Equation
1 Mark
$$m = 5 - 3 = 2$$
Check: LHS $= 5^2 \div 5^{-3} = 5^{2 - (-3)} = 5^5 = \text{RHS}$ (Verified).
$$\mathbf{\text{Answer: } m = 2}$$
Check: LHS $= 5^2 \div 5^{-3} = 5^{2 - (-3)} = 5^5 = \text{RHS}$ (Verified).
$$\mathbf{\text{Answer: } m = 2}$$
Examiner Scoring Key: Explicitly stating 'Since bases are equal, exponents are equal' carries 1 mark. Sign error in m - (-3) = m + 3 forfeits 2 marks.
Model 3: Prime Harmonization of Multi-Term Expression
7 Marks
Simplify the following expression, giving the final answer in terms of powers of prime factors:
$$\frac{25 \times 5^2 \times t^8}{10^3 \times t^4} \quad (t
eq 0)$$
$$\frac{25 \times 5^2 \times t^8}{10^3 \times t^4} \quad (t
eq 0)$$
Step-by-Step Model Solution:
Step 1: Express Composite Bases as Prime Factors
2 Marks
Identify composite numbers: $25$ and $10$.
$25 = 5^2$
$10 = 2 \times 5 \implies 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$ (using $(ab)^n = a^n b^n$).
Substitute these prime decompositions into the expression:
$$= \frac{5^2 \times 5^2 \times t^8}{2^3 \times 5^3 \times t^4}$$
$25 = 5^2$
$10 = 2 \times 5 \implies 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$ (using $(ab)^n = a^n b^n$).
Substitute these prime decompositions into the expression:
$$= \frac{5^2 \times 5^2 \times t^8}{2^3 \times 5^3 \times t^4}$$
Step 2: Combine Powers with Same Base in Numerator
2 Marks
Using $a^m \times a^n = a^{m+n}$ for base $5$:
$$\text{Numerator} = 5^{2+2} \times t^8 = 5^4 \times t^8$$
So the expression becomes:
$$= \frac{5^4 \times t^8}{2^3 \times 5^3 \times t^4}$$
$$\text{Numerator} = 5^{2+2} \times t^8 = 5^4 \times t^8$$
So the expression becomes:
$$= \frac{5^4 \times t^8}{2^3 \times 5^3 \times t^4}$$
Step 3: Apply Quotient Law to Reduce Powers
3 Marks
Using $\frac{a^m}{a^n} = a^{m-n}$ for base $5$ and base $t$:
$$= \frac{5^{4-3} \times t^{8-4}}{2^3} = \frac{5^1 \times t^4}{2^3} = \frac{5 t^4}{8}$$
$$\mathbf{\text{Answer: } \frac{5}{8}t^4}$$
$$= \frac{5^{4-3} \times t^{8-4}}{2^3} = \frac{5^1 \times t^4}{2^3} = \frac{5 t^4}{8}$$
$$\mathbf{\text{Answer: } \frac{5}{8}t^4}$$
Examiner Scoring Key: Prime factor decomposition carries 2 marks. Clean cancellation using quotient law carries 3 marks.
Model 4: Negative Indices with Composite Bases
7 Marks
Simplify the expression by eliminating negative exponents and factorizing into prime factors:
$$\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$
$$\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$
Step-by-Step Model Solution:
Step 1: Convert Negative Exponents to Positive Powers
2 Marks
Using $a^{-n} = \frac{1}{a^n}$ and $\frac{1}{a^{-n}} = a^n$:
Move terms with negative powers from numerator to denominator and vice versa:
$$= \frac{5^7 \times 6^5 \times 125}{3^5 \times 10^5}$$
Move terms with negative powers from numerator to denominator and vice versa:
$$= \frac{5^7 \times 6^5 \times 125}{3^5 \times 10^5}$$
Step 2: Prime Factorize All Composite Bases
2 Marks
Decompose bases into primes:
$125 = 5^3$
$6^5 = (2 \times 3)^5 = 2^5 \times 3^5$
$10^5 = (2 \times 5)^5 = 2^5 \times 5^5$
Substitute back into the expression:
$$= \frac{5^7 \times (2^5 \times 3^5) \times 5^3}{3^5 \times (2^5 \times 5^5)}$$
$125 = 5^3$
$6^5 = (2 \times 3)^5 = 2^5 \times 3^5$
$10^5 = (2 \times 5)^5 = 2^5 \times 5^5$
Substitute back into the expression:
$$= \frac{5^7 \times (2^5 \times 3^5) \times 5^3}{3^5 \times (2^5 \times 5^5)}$$
Step 3: Combine Like Bases and Cancel
3 Marks
Numerator: $2^5 \times 3^5 \times 5^{7+3} = 2^5 \times 3^5 \times 5^{10}$.
Denominator: $2^5 \times 3^5 \times 5^5$.
Using the Quotient Law:
$$= 2^{5-5} \times 3^{5-5} \times 5^{10-5} = 2^0 \times 3^0 \times 5^5$$
Since $2^0 = 1$ and $3^0 = 1$:
$$= 1 \times 1 \times 5^5 = 5^5 = 3125$$
$$\mathbf{\text{Answer: } 5^5 \text{ or } 3125}$$
Denominator: $2^5 \times 3^5 \times 5^5$.
Using the Quotient Law:
$$= 2^{5-5} \times 3^{5-5} \times 5^{10-5} = 2^0 \times 3^0 \times 5^5$$
Since $2^0 = 1$ and $3^0 = 1$:
$$= 1 \times 1 \times 5^5 = 5^5 = 3125$$
$$\mathbf{\text{Answer: } 5^5 \text{ or } 3125}$$
Examiner Scoring Key: Award 2 marks for reciprocating negative powers, 2 marks for prime factoring 6, 10, 125, and 3 marks for final reduction to 5^5 = 3125.
Model 5: Advanced Exponential Equation
10 Marks
Solve the following exponential equation for the unknown variable $x$:
$$2^{x-1} + 2^{x+1} = 320$$
$$2^{x-1} + 2^{x+1} = 320$$
Step-by-Step Model Solution:
Step 1: Decompose Terms using Product Law of Exponents
3 Marks
We know that $a^{m+n} = a^m \times a^n$.
Rewrite $2^{x-1}$ and $2^{x+1}$ in terms of $2^x$:
$$2^{x-1} = 2^x \times 2^{-1} = 2^x \times \frac{1}{2}$$
$$2^{x+1} = 2^x \times 2^1 = 2^x \times 2$$
Substitute these into the equation:
$$2^x \times \frac{1}{2} + 2^x \times 2 = 320$$
Rewrite $2^{x-1}$ and $2^{x+1}$ in terms of $2^x$:
$$2^{x-1} = 2^x \times 2^{-1} = 2^x \times \frac{1}{2}$$
$$2^{x+1} = 2^x \times 2^1 = 2^x \times 2$$
Substitute these into the equation:
$$2^x \times \frac{1}{2} + 2^x \times 2 = 320$$
Step 2: Factor Out Common Term $2^x$
3 Marks
Take $2^x$ common from both terms on the LHS:
$$2^x \left( \frac{1}{2} + 2 \right) = 320$$
Simplify inside the bracket:
$$\frac{1}{2} + 2 = \frac{1 + 4}{2} = \frac{5}{2}$$
So the equation becomes:
$$2^x \times \frac{5}{2} = 320$$
$$2^x \left( \frac{1}{2} + 2 \right) = 320$$
Simplify inside the bracket:
$$\frac{1}{2} + 2 = \frac{1 + 4}{2} = \frac{5}{2}$$
So the equation becomes:
$$2^x \times \frac{5}{2} = 320$$
Step 3: Isolate $2^x$ and Factorize RHS
2 Marks
Multiply both sides by $\frac{2}{5}$:
$$2^x = 320 \times \frac{2}{5} = \frac{320}{5} \times 2 = 64 \times 2 = 128$$
Now express $128$ as a power of base $2$:
$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.
$$2^x = 320 \times \frac{2}{5} = \frac{320}{5} \times 2 = 64 \times 2 = 128$$
Now express $128$ as a power of base $2$:
$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.
Step 4: Equate Exponents & Verify
2 Marks
$$2^x = 2^7$$
Since bases are equal ($2 = 2$), equate exponents:
$$x = 7$$
Check: $2^{7-1} + 2^{7+1} = 2^6 + 2^8 = 64 + 256 = 320 = \text{RHS}$.
$$\mathbf{\text{Answer: } x = 7}$$
Since bases are equal ($2 = 2$), equate exponents:
$$x = 7$$
Check: $2^{7-1} + 2^{7+1} = 2^6 + 2^8 = 64 + 256 = 320 = \text{RHS}$.
$$\mathbf{\text{Answer: } x = 7}$$
Examiner Scoring Key: Factoring out 2^x is the crucial algebraic step carrying 3 marks. Verification is recommended for full 10 marks.
Model 6: Scientific Notation & Astronomical Ratios
10 Marks
The mass of the Earth is estimated to be $5.97 \times 10^{24}\text{ kg}$ and that of the Moon is $7.35 \times 10^{22}\text{ kg}$.
(a) What is the combined total mass of Earth and Moon in standard scientific notation?
(b) Approximately how many times heavier is the Earth compared to the Moon?
(a) What is the combined total mass of Earth and Moon in standard scientific notation?
(b) Approximately how many times heavier is the Earth compared to the Moon?
Step-by-Step Model Solution:
Step 1: Harmonize Powers of 10 for Addition
3 Marks
Mass of Earth $M_E = 5.97 \times 10^{24}\text{ kg}$.
Mass of Moon $M_M = 7.35 \times 10^{22}\text{ kg}$.
To add these two quantities, convert both to the same exponent of $10$ (use $10^{24}$ or $10^{22}$):
$$M_M = 7.35 \times 10^{22} = \frac{7.35}{10^2} \times 10^{24} = 0.0735 \times 10^{24}\text{ kg}$$
Mass of Moon $M_M = 7.35 \times 10^{22}\text{ kg}$.
To add these two quantities, convert both to the same exponent of $10$ (use $10^{24}$ or $10^{22}$):
$$M_M = 7.35 \times 10^{22} = \frac{7.35}{10^2} \times 10^{24} = 0.0735 \times 10^{24}\text{ kg}$$
Step 2: Calculate Combined Total Mass (Part a)
3 Marks
$$\text{Total Mass } M_{\text{total}} = M_E + M_M = (5.97 \times 10^{24}) + (0.0735 \times 10^{24})$$
$$= (5.97 + 0.0735) \times 10^{24} = 6.0435 \times 10^{24}\text{ kg}$$
Since $1.0 \le 6.0435 < 10$, this is in correct standard scientific form:
$$\mathbf{M_{\text{total}} = 6.0435 \times 10^{24}\text{ kg}}$$
$$= (5.97 + 0.0735) \times 10^{24} = 6.0435 \times 10^{24}\text{ kg}$$
Since $1.0 \le 6.0435 < 10$, this is in correct standard scientific form:
$$\mathbf{M_{\text{total}} = 6.0435 \times 10^{24}\text{ kg}}$$
Step 3: Calculate Ratio of Masses (Part b)
3 Marks
$$\text{Ratio } R = \frac{M_E}{M_M} = \frac{5.97 \times 10^{24}}{7.35 \times 10^{22}}$$
Using the Quotient Law for powers of $10$:
$$\frac{10^{24}}{10^{22}} = 10^{24-22} = 10^2 = 100$$
$$R = \frac{5.97}{7.35} \times 100 = \frac{597}{7.35} \approx 81.22$$
Using the Quotient Law for powers of $10$:
$$\frac{10^{24}}{10^{22}} = 10^{24-22} = 10^2 = 100$$
$$R = \frac{5.97}{7.35} \times 100 = \frac{597}{7.35} \approx 81.22$$
Step 4: Formal Concluding Statement
1 Mark
1. The combined mass of Earth and Moon is $\mathbf{6.0435 \times 10^{24}\text{ kg}}$.
2. The Earth is approximately $\mathbf{81.2\text{ times}}$ (or roughly $81$ times) heavier than the Moon.
2. The Earth is approximately $\mathbf{81.2\text{ times}}$ (or roughly $81$ times) heavier than the Moon.
Examiner Scoring Key: In part (a), converting 7.35 * 10^22 to 0.0735 * 10^24 before adding is required. In part (b), index subtraction 10^(24-22) = 10^2 carries 2 marks.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
Evaluate the following numeric expression showing complete step-by-step reduction:
$$\left(3^0 + 4^{-1}\right) \times 2^2$$
$$\left(3^0 + 4^{-1}\right) \times 2^2$$
Model Descriptive Solution:
Step 1: Evaluate Terms Inside Parentheses
2.5 Marks
Using the Zero Exponent Law: $3^0 = 1$.
Using the Negative Exponent Law: $4^{-1} = \frac{1}{4^1} = \frac{1}{4}$.
Now add the fractions inside the bracket:
$$3^0 + 4^{-1} = 1 + \frac{1}{4} = \frac{4 + 1}{4} = \frac{5}{4}$$
Using the Negative Exponent Law: $4^{-1} = \frac{1}{4^1} = \frac{1}{4}$.
Now add the fractions inside the bracket:
$$3^0 + 4^{-1} = 1 + \frac{1}{4} = \frac{4 + 1}{4} = \frac{5}{4}$$
Step 2: Multiply by $2^2$
2.5 Marks
Evaluate $2^2 = 4$.
$$\left(\frac{5}{4}\right) \times 4 = 5$$
$$\left(\frac{5}{4}\right) \times 4 = 5$$
Final Answer: $5$
Question 2
5 Marks
Express the following numbers in standard scientific notation:
(a) $0.000000000085$
(b) $602,000,000,000,000,000$
(a) $0.000000000085$
(b) $602,000,000,000,000,000$
Model Descriptive Solution:
Step 1: Part (a) Conversion
2.5 Marks
In $0.000000000085$, shift the decimal point $11$ places to the right to place it after the first non-zero digit $8$:
$$0.000000000085 = 8.5 \times 10^{-11}$$
$$0.000000000085 = 8.5 \times 10^{-11}$$
Step 2: Part (b) Conversion
2.5 Marks
In $602,000,000,000,000,000$, shift the decimal point $17$ places to the left to place it after $6$:
$$602,000,000,000,000,000 = 6.02 \times 10^{17}$$
$$602,000,000,000,000,000 = 6.02 \times 10^{17}$$
Final Answer: (a) $8.5 \times 10^{-11}$; (b) $6.02 \times 10^{17}$
Question 3
5 Marks
Find the value of $m$ if:
$$(-3)^{m+1} \times (-3)^5 = (-3)^7$$
$$(-3)^{m+1} \times (-3)^5 = (-3)^7$$
Model Descriptive Solution:
Step 1: Combine Powers on LHS
2.5 Marks
By the Product Law: $a^p \times a^q = a^{p+q}$ where base $a = -3$:
$$(-3)^{(m+1) + 5} = (-3)^7 \implies (-3)^{m+6} = (-3)^7$$
$$(-3)^{(m+1) + 5} = (-3)^7 \implies (-3)^{m+6} = (-3)^7$$
Step 2: Equate Exponents
2.5 Marks
Since the bases on both sides are identical $(-3)$, equate exponents:
$$m + 6 = 7 \implies m = 7 - 6 = 1$$
$$m + 6 = 7 \implies m = 7 - 6 = 1$$
Final Answer: $m = 1$
Question 4
7 Marks
Simplify the following expression using laws of indices:
$$\left\{ \left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3} \right\} \div \left(\frac{1}{4}\right)^{-2}$$
$$\left\{ \left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3} \right\} \div \left(\frac{1}{4}\right)^{-2}$$
Model Descriptive Solution:
Step 1: Invert Negative Powers Inside Braces
3 Marks
Using $\left(\frac{1}{a}\right)^{-n} = a^n$:
$$\left(\frac{1}{3}\right)^{-2} = 3^2 = 9$$
$$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$$
Inner expression $= 9 - 8 = 1$.
$$\left(\frac{1}{3}\right)^{-2} = 3^2 = 9$$
$$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$$
Inner expression $= 9 - 8 = 1$.
Step 2: Evaluate Divisor
2 Marks
$$\left(\frac{1}{4}\right)^{-2} = 4^2 = 16$$
Step 3: Perform Division
2 Marks
$$1 \div 16 = \frac{1}{16}$$
Final Answer: $\frac{1}{16}$
Question 5
7 Marks
By what rational number should $(-4)^{-1}$ be multiplied so that the resulting product becomes $10^{-1}$?
Model Descriptive Solution:
Step 1: Set Up Algebraic Equation
2 Marks
Let the required rational multiplier be $x$.
According to the question:
$$(-4)^{-1} \times x = 10^{-1}$$
According to the question:
$$(-4)^{-1} \times x = 10^{-1}$$
Step 2: Convert Negative Exponents to Fractions
2 Marks
$$(-4)^{-1} = -\frac{1}{4}, \quad 10^{-1} = \frac{1}{10}$$
$$-\frac{1}{4} \times x = \frac{1}{10}$$
$$-\frac{1}{4} \times x = \frac{1}{10}$$
Step 3: Solve for x
3 Marks
$$x = \frac{1}{10} \div \left(-\frac{1}{4}\right) = \frac{1}{10} \times (-4) = -\frac{4}{10} = -\frac{2}{5}$$
Final Answer: $-\frac{2}{5}$
Question 6
7 Marks
Simplify the following expression and state the result as an irreducible fraction:
$$\frac{2^{-1} \times 4^{-1}}{2^{-2}}$$
$$\frac{2^{-1} \times 4^{-1}}{2^{-2}}$$
Model Descriptive Solution:
Step 1: Convert All Bases to Prime Base 2
3 Marks
Notice that $4 = 2^2$. Therefore:
$$4^{-1} = (2^2)^{-1} = 2^{2 \times (-1)} = 2^{-2}$$
Numerator $= 2^{-1} \times 2^{-2} = 2^{-1 + (-2)} = 2^{-3}$.
$$4^{-1} = (2^2)^{-1} = 2^{2 \times (-1)} = 2^{-2}$$
Numerator $= 2^{-1} \times 2^{-2} = 2^{-1 + (-2)} = 2^{-3}$.
Step 2: Apply Quotient Law
2 Marks
$$\frac{2^{-3}}{2^{-2}} = 2^{-3 - (-2)} = 2^{-3 + 2} = 2^{-1}$$
Step 3: Convert to Fraction
2 Marks
$$2^{-1} = \frac{1}{2}$$
Final Answer: $\frac{1}{2}$
Question 7
7 Marks
Solve the following exponential equation for $x$:
$$3^{2x-1} = \frac{1}{27^{x-3}}$$
$$3^{2x-1} = \frac{1}{27^{x-3}}$$
Model Descriptive Solution:
Step 1: Express RHS with Base 3
3 Marks
We know $27 = 3^3$. Therefore:
$$27^{x-3} = (3^3)^{x-3} = 3^{3(x-3)} = 3^{3x-9}$$
Now rewrite the fraction using negative exponents:
$$\frac{1}{27^{x-3}} = \frac{1}{3^{3x-9}} = 3^{-(3x-9)} = 3^{-3x+9}$$
$$27^{x-3} = (3^3)^{x-3} = 3^{3(x-3)} = 3^{3x-9}$$
Now rewrite the fraction using negative exponents:
$$\frac{1}{27^{x-3}} = \frac{1}{3^{3x-9}} = 3^{-(3x-9)} = 3^{-3x+9}$$
Step 2: Equate Exponents of Equal Bases
2 Marks
The equation is now: $3^{2x-1} = 3^{-3x+9}$.
Since bases are equal ($3 = 3$):
$$2x - 1 = -3x + 9$$
Since bases are equal ($3 = 3$):
$$2x - 1 = -3x + 9$$
Step 3: Solve Linear Equation
2 Marks
$$2x + 3x = 9 + 1 \implies 5x = 10 \implies x = \frac{10}{5} = 2$$
Final Answer: $x = 2$
Question 8
10 Marks
Simplify the following expression by converting all bases into prime factors:
$$\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$$
$$\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$$
Model Descriptive Solution:
Step 1: Prime Factorize All Bases
3 Marks
$12 = 2^2 \times 3 \implies 12^4 = (2^2 \times 3)^4 = 2^8 \times 3^4$
$9 = 3^2 \implies 9^3 = (3^2)^3 = 3^6$
$4 = 2^2$
$6 = 2 \times 3 \implies 6^3 = 2^3 \times 3^3$
$8 = 2^3 \implies 8^2 = (2^3)^2 = 2^6$
$27 = 3^3$
$9 = 3^2 \implies 9^3 = (3^2)^3 = 3^6$
$4 = 2^2$
$6 = 2 \times 3 \implies 6^3 = 2^3 \times 3^3$
$8 = 2^3 \implies 8^2 = (2^3)^2 = 2^6$
$27 = 3^3$
Step 2: Combine Powers of 2 and 3 in Numerator and Denominator
3 Marks
Numerator:
$$2^8 \times 3^4 \times 3^6 \times 2^2 = 2^{8+2} \times 3^{4+6} = 2^{10} \times 3^{10}$$
Denominator:
$$(2^3 \times 3^3) \times 2^6 \times 3^3 = 2^{3+6} \times 3^{3+3} = 2^9 \times 3^6$$
$$2^8 \times 3^4 \times 3^6 \times 2^2 = 2^{8+2} \times 3^{4+6} = 2^{10} \times 3^{10}$$
Denominator:
$$(2^3 \times 3^3) \times 2^6 \times 3^3 = 2^{3+6} \times 3^{3+3} = 2^9 \times 3^6$$
Step 3: Apply Quotient Law & Calculate Final Value
4 Marks
$$= 2^{10-9} \times 3^{10-6} = 2^1 \times 3^4$$
Since $3^4 = 81$:
$$= 2 \times 81 = 162$$
Since $3^4 = 81$:
$$= 2 \times 81 = 162$$
Final Answer: $162$
Question 9
10 Marks
If $5^{x-3} \times 3^{2x-8} = 225$, determine the value of $x$. Show complete prime factorization working.
Model Descriptive Solution:
Step 1: Prime Factorize RHS Number 225
3 Marks
Perform prime factorization of $225$:
$225 = 5 \times 45 = 5 \times 5 \times 9 = 5^2 \times 3^2$.
The equation becomes:
$$5^{x-3} \times 3^{2x-8} = 5^2 \times 3^2$$
$225 = 5 \times 45 = 5 \times 5 \times 9 = 5^2 \times 3^2$.
The equation becomes:
$$5^{x-3} \times 3^{2x-8} = 5^2 \times 3^2$$
Step 2: Equate Corresponding Exponents of Prime Bases
4 Marks
Since bases $5$ and $3$ are distinct prime numbers, the powers on both sides must match independently:
Equating exponents of base $5$:
$$x - 3 = 2 \implies x = 2 + 3 = 5$$
Equating exponents of base $3$:
$$2x - 8 = 2 \implies 2x = 10 \implies x = 5$$
Equating exponents of base $5$:
$$x - 3 = 2 \implies x = 2 + 3 = 5$$
Equating exponents of base $3$:
$$2x - 8 = 2 \implies 2x = 10 \implies x = 5$$
Step 3: Consistency Check & Conclusion
3 Marks
Both independent prime base equations yield the consistent unique solution $x = 5$.
Check: $5^{5-3} \times 3^{2(5)-8} = 5^2 \times 3^2 = 25 \times 9 = 225$ (Verified!).
Check: $5^{5-3} \times 3^{2(5)-8} = 5^2 \times 3^2 = 25 \times 9 = 225$ (Verified!).
Final Answer: $x = 5$
Question 10
10 Marks
In a library archive, a stack contains $5$ books each of thickness $20\text{ mm}$ and $5$ sheets of parchment paper each of thickness $0.016\text{ mm}$. Calculate the total thickness of the stack in millimetres, and express this total in standard scientific notation.
Model Descriptive Solution:
Step 1: Calculate Total Thickness of Books
2.5 Marks
Number of books $= 5$. Thickness of 1 book $= 20\text{ mm}$.
$$\text{Total thickness of books} = 5 \times 20 = 100\text{ mm}$$
$$\text{Total thickness of books} = 5 \times 20 = 100\text{ mm}$$
Step 2: Calculate Total Thickness of Paper Sheets
2.5 Marks
Number of sheets $= 5$. Thickness of 1 sheet $= 0.016\text{ mm}$.
$$\text{Total thickness of sheets} = 5 \times 0.016 = 0.080\text{ mm} = 0.08\text{ mm}$$
$$\text{Total thickness of sheets} = 5 \times 0.016 = 0.080\text{ mm} = 0.08\text{ mm}$$
Step 3: Sum the Thicknesses
2 Marks
$$\text{Total Stack Thickness} = 100\text{ mm} + 0.08\text{ mm} = 100.08\text{ mm}$$
Step 4: Convert to Standard Scientific Notation
3 Marks
Shift decimal point $2$ places to the left to obtain $1.0008$ (where $1 \le 1.0008 < 10$):
$$100.08 = 1.0008 \times 10^2\text{ mm}$$
In centimetres: $1.0008 \times 10^1\text{ cm}$.
$$100.08 = 1.0008 \times 10^2\text{ mm}$$
In centimetres: $1.0008 \times 10^1\text{ cm}$.
Final Answer: $1.0008 \times 10^2\text{ mm}$
High-Yield FAQs & Revision Notes
What is the difference between (-2)^4 and -2^4?
In $(-2)^4$, the negative sign is part of the base, so $(-2) \times (-2) \times (-2) \times (-2) = +16$. In $-2^4$, the exponent applies only to $2$, so $-(2 \times 2 \times 2 \times 2) = -16$.
Why is any non-zero number to the power 0 equal to 1?
By the quotient rule, $\frac{a^m}{a^m} = a^{m-m} = a^0$. Any non-zero number divided by itself equals $1$, hence $a^0 = 1$.
What is 0^0?
In elementary mathematics and arithmetic, $0^0$ is an indeterminate form and is undefined.
How do you handle negative exponents on fractions?
Simply invert the fraction and change the negative exponent to positive: $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$.
What is standard scientific notation?
Standard scientific form expresses any number as $m \times 10^k$, where $1.0 \le m < 10$ and $k$ is an integer.
How do you solve exponential equations where bases are different?
Decompose both sides into prime factor bases (e.g., $2, 3, 5$). Equate the exponents of corresponding prime bases independently.
Can exponents be fractional in RIMC?
Yes. Simple fractional exponents represent roots: $a^{1/2} = \sqrt{a}$, $a^{1/3} = \sqrt[3]{a}$, and $a^{m/n} = (\sqrt[n]{a})^m$.
Why should composite bases be prime factorized first?
Converting composite bases like $4, 6, 8, 9, 12$ into prime powers ($2^2, 2 \times 3, 2^3, 3^2, 2^2 \times 3$) allows combining terms using the standard laws of indices.
What is the order of operations when evaluating 2^3^2?
Expressions in towers of powers without parentheses are evaluated from the top down: $2^{(3^2)} = 2^9 = 512$, whereas $(2^3)^2 = 2^6 = 64$.
How does this module map to the official RIMC syllabus?
Directly covers the 'Exponents and indices (simple problem)' subpart of Serial No. 46 of Chapter 8 broader syllabus in the RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. Part of 46 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Product Law | $a^m \times a^n = a^{m+n}$. Add powers when bases are identical. |
| Quotient Law | $a^m \div a^n = a^{m-n}$. Subtract denominator power from numerator. |
| Power of a Power | $(a^m)^n = a^{mn}$. Multiply inner and outer powers. |
| Negative Power | $a^{-n} = 1/a^n$ and $(a/b)^{-n} = (b/a)^n$. Invert base to make power positive. |
| Zero Exponent | $a^0 = 1$ for all $a \neq 0$. |
| Scientific Form | $m \times 10^k$ where $1.0 \le m < 10$. Exactly one non-zero digit before decimal point. |