For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: integer and fractional indices, index laws, standard form, estimation, and magnitude checks form the main route.
School-sensitive extension: more demanding proof, scientific-context modelling, and links to logarithms may be introduced earlier in some schools.
2027 national comparison: K310 Topic N1 includes standard form, positive, negative, zero and fractional indices, approximation, estimation, and calculator work alongside earlier number foundations.
Check your school: confirm significant-figure conventions, calculator display expectations, and the depth of scientific applications.
Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 02) Indices and Standard Form cover? A: Index laws and scientific notation techniques for fast, accurate manipulation of very large or small quantities.
Scientific notation keeps working memory clear when comparing magnitudes or evaluating calculator steps. Pair it with consistent index laws.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Indices are rules for handling repeated multiplication.
Use it as a working check: For the same base, multiply by adding powers, divide by subtracting powers, and raise a power to a power by multiplying powers.
Then go one layer deeper: Example: in standard form, multiply the front numbers and combine the powers of ten, then adjust so the front number is at least 1 and less than 10.
Essential identities
am×an=am+n and anam=am−n for a=0.
(am)n=amn and a−n=an1
Standard form: N=k×10n with 1≤k<10 and n
Standard-form adjustment checkpoint
After multiplying or dividing standard-form numbers, the first number may no longer be between 1 and 10. Fix the front number and compensate the power of 10 in the opposite direction.
Starting product
Front-number move
Power-of-ten move
Standard form
12.6×102
Move decimal 1 place left: 12.6=1.26×101.
Increase the power by 1.
1.26×103
0.75×1012
Move decimal 1 place right: 0.75=7.5×10−1.
Worked check: (6×104)(8×10−2)=48×102. Since 48 is too large for standard form, write 48=4.8×101, so the final answer is 4.8×103.
Misconception check: moving the decimal left makes the front number smaller, so the power of 10 must increase to keep the original value unchanged.
Negative-index position checkpoint
When variables appear in a fraction, decide the index first and the final position second. Do not move a term across the fraction bar until the exponent for that base has been simplified.
Situation
First move
Then write with positive indices
Common trap
Same base multiplied
Add the indices.
Move the base only if the final index is negative.
Moving every negative-index term immediately and losing like bases.
Same base divided
Subtract the denominator index.
If the final index is negative, put that base in the denominator.
Treating m−(−n) as m−n.
Bracket with an outside power
Multiply every index inside the bracket by the outside power.
Simplify all like bases before moving positions.
Applying the outside power to the coefficient but not to each index.
Worked check: for x−5x−2, subtract the denominator index first:
x−2−(−5)=x3.
The answer is x3, not x71. Misconception check: dividing by x−5 is not the same as subtracting 5; the denominator index is negative, so subtracting it increases the final index.
Worked example - Multiply and compare magnitudes
Compute (4.2×105)(3×10−3) and express the answer in standard form.
Multiply the decimal parts: 4.2×3=12.6.
Add the indices: 105×10−3=102.
Combine: 12.6×102.
Adjust to standard form: 12.6=1.26×101, so 12.6×102=1.26×103
Hence the product is 1.26×103.
Worked example - Simplify negative indices with variables
Simplify 10x−1y2(2x−3y4)2×5x2y−1 and express the final answer with positive indices only.
Expand the bracket: (2x−3y4)2=4x−6y8.
Multiply by 5x2y−1: the numerator becomes 20x−4y7 (add exponents: −6+2=−4
Divide by 10x−1y2: coefficients give 20÷10=2; subtract indices for like bases so x−4−(−1)=x−3
Rewrite with positive indices: 2x−3y5=x32y5
Therefore the simplified expression is x32y5.
Worked example - Use standard form in a real-world estimate
A lab sample has a total mass of 2.4×10−3 grams and each bacterium has mass 3.2×10−15 grams. Estimate the number of bacteria present, giving the answer in standard form.
Model the count as 3.2×10−152.4×10−3.
Divide the decimal parts: 2.4÷3.2=0.75.
Subtract the powers of ten: 10−3÷10−15=1012.
Combine: 0.75×1012.
Adjust to standard form (leading digit between 1 and 10): 0.75=7.5×10−1, so 0.75×1012=7.5×1011
The sample contains approximately 7.5×1011 bacteria.
Practice Quiz
Check your recall of index rules and standard form conversions with quick-fire questions.
Try this
Write 2.1×10−40.00084 in standard form, showing every index step.
