Unit Synthesis — Index Laws Review
All six index laws on one page, plus scientific notation and fractional indices. Mixed exam-style problems that combine everything you've learned in Unit 1.
Printable Worksheets
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List every index law you can remember from L01–L19. Try to write each one as a formula. Don't peek — you'll check your list against the next card.
Here are all six index laws covered in Unit 1. Memorise these cold — every question in this unit reduces to combining them.
1. Product: $x^m \cdot x^n = x^{m+n}$. 2. Quotient: $\dfrac{x^m}{x^n} = x^{m-n}$. 3. Power of a power: $(x^m)^n = x^{mn}$. 4. Power of a product: $(xy)^n = x^n y^n$. 5. Zero index: $x^0 = 1$ ($x \ne 0$). 6. Negative index: $x^{-n} = \dfrac{1}{x^n}$. Bonus: $x^{m/n} = \sqrt[n]{x^m}$.
Know
- All six index laws (and the fractional bonus law)
- Standard form $a \times 10^n$ with $1 \le a < 10$
- Convention: most answers to 3 sig fig
Understand
- How to chain laws together in one expression
- When “root first” saves you arithmetic
- Why scientific notation is a special case of the same laws
Can Do
- Simplify mixed integer-and-fractional-index expressions
- Calculate with scientific notation accurately
- Solve word problems involving cosmic/atomic scales
Wrong: $(2 a)^3 = 2 a^3$ — forgetting the coefficient takes the power.
Right: $(2 a)^3 = 2^3 a^3 = 8 a^3$.
Wrong: $x^{-2} = -x^2$ — treating the negative as a sign on the base.
Right: $x^{-2} = \dfrac{1}{x^2}$. Negative index flips the base.
Wrong: Leaving a scientific-notation coefficient outside $[1, 10)$, e.g. $12 \times 10^9$.
Right: Re-normalise: $12 \times 10^9 = 1.2 \times 10^{10}$.
Use this map to revise. Each row links a lesson to its key formula.
L01–L05: base, exponent, product/quotient/power laws — numbers. L06–L09: zero index, negative index, and converting between fractional and negative forms. L10–L13: applying laws to algebraic expressions, multi-variable terms. L14–L15: scientific notation for large/small numbers, converting to/from standard form. L16–L17: arithmetic in scientific notation, comparing orders of magnitude. L18: applications, sig fig, calculator. L19: fractional indices and surds.
Simplify $\dfrac{(3 x^2)^2 \cdot x^{-3}}{9 x^0}$. We will need: power-of-a-product, product, negative index, zero index, and quotient.
Power first: $(3 x^2)^2 = 9 x^4$. Zero: $x^0 = 1$. Negative: $x^{-3} = \dfrac{1}{x^3}$, but keep as $x^{-3}$ for the product step. Product on numerator: $9 x^4 \cdot x^{-3} = 9 x^{4-3} = 9 x^1$. Quotient: $\dfrac{9 x}{9 \cdot 1} = x$.
Watch Me Solve It · 3 mixed examples
- 1Divide coefficients$6 / 2 = 3$
- 2Quotient rule on each variable$a^{5-2} = a^3$; $\;b^{-2 - (-5)} = b^{3}$
- 3Combine$3 a^3 b^3$
- 1Set up the product$N = (4.0 \times 10^9) \times (2.5 \times 10^3)$
- 2Multiply coefficients, add indices$4.0 \times 2.5 = 10$; $\;10^{9+3} = 10^{12}$
- 3Re-normalise to standard form (2 s.f.)$10 \times 10^{12} = 1.0 \times 10^{13}$
- 1Power of a product on the bracket$(2 a^3)^2 = 4 a^6$
- 2Product rule on the numerator$4 a^6 \cdot a^{-1} = 4 a^{6 - 1} = 4 a^5$
- 3Quotient rule with fractional index$\dfrac{4 a^5}{4 a^{1/2}} = a^{5 - 1/2} = a^{10/2 - 1/2} = a^{9/2}$
Top Mistakes To Avoid In The Exam
Integer indices
- Product: $x^m \cdot x^n = x^{m+n}$
- Quotient: $x^m / x^n = x^{m-n}$
- Power: $(x^m)^n = x^{mn}$
- Of a product: $(xy)^n = x^n y^n$
Zero & negative
- $x^0 = 1$ ($x \ne 0$)
- $x^{-n} = \dfrac{1}{x^n}$
- $\left(\dfrac{x}{y}\right)^{-n} = \left(\dfrac{y}{x}\right)^n$
Fractional & scientific
- $x^{1/n} = \sqrt[n]{x}$
- $x^{m/n} = \sqrt[n]{x^m}$
- $a \times 10^n$, $1 \le a < 10$
Order of operations
- Brackets / powers $\to$ multiply $\to$ divide
- Re-normalise scientific notation at the end
- Final check: positive index? right sig fig?
How are you completing this lesson?
Brain Trainer · 5 mixed drills
One question from each strand: integer indices, zero/negative, algebraic, scientific notation, fractional.
1 Simplify $x^7 \cdot x^{-3}$ as a positive index.
$x^{7 + (-3)} = x^4$.$x^4$2 Evaluate $5^0 + 5^{-1}$.
$5^0 = 1$; $5^{-1} = 1/5 = 0.2$; sum $= 1.2$.$1.2$ or $\dfrac{6}{5}$3 Simplify $\dfrac{(3 x^2 y)^2}{9 x y}$.
Numerator $= 9 x^4 y^2$; quotient $= x^{4-1} y^{2-1} = x^3 y$.$x^3 y$4 Calculate $(2.5 \times 10^4) \times (4 \times 10^{-7})$.
$2.5 \times 4 = 10$; $10^{4-7} = 10^{-3}$; re-normalise $10 \times 10^{-3} = 1 \times 10^{-2}$.$1 \times 10^{-2}$ or $0.01$5 Evaluate $16^{3/4}$.
$\sqrt[4]{16} = 2$; $2^3 = 8$.$8$
Quick Check · 5 questions (mixed)
Show Your Working · 3 exam-style questions
Q6. Simplify, expressing each answer with positive indices only: (a) $a^5 \cdot a^{-2}$, (b) $(2 x^3 y)^4$, (c) $\dfrac{12 m^2 n^{-1}}{4 m^{-3} n^2}$.
Q7. A red blood cell has a volume of $9.0 \times 10^{-14}$ L. An adult has about $5.0$ L of blood, of which $45\%$ is red blood cells. Estimate the number of red blood cells in their body, to 2 sig fig.
Q8. (a) Simplify $\dfrac{(2 x^2)^3 \cdot x^{-1}}{4 x^{1/2}}$, stating each index law used. (b) Hence evaluate when $x = 4$.
Quick Check
1. D — $x^5$.
2. B — $1$.
3. A — $27 x^6$.
4. C — $2.4 \times 10^4$.
5. D — $25$.
Show Your Working Model Answers
Q6 (3 marks): (a) $a^{5 + (-2)} = a^3$ [1]; (b) $(2 x^3 y)^4 = 2^4 \cdot x^{12} \cdot y^4 = 16 x^{12} y^4$ [1]; (c) $\dfrac{12}{4} = 3$; $m^{2 - (-3)} = m^5$; $n^{-1 - 2} = n^{-3}$ — answer with positive indices: $\dfrac{3 m^5}{n^3}$ [1].
Q7 (3 marks): Volume of RBCs = $5.0 \times 0.45 = 2.25$ L [1]. Number $= \dfrac{2.25}{9.0 \times 10^{-14}} = \dfrac{2.25}{9.0} \times 10^{14} = 0.25 \times 10^{14}$ [1] $= 2.5 \times 10^{13}$ red blood cells (2 s.f.) [1].
Q8 (4 marks): (a) Power of a product: $(2 x^2)^3 = 8 x^6$ [1]. Product on numerator: $8 x^6 \cdot x^{-1} = 8 x^5$ [1]. Quotient with fractional index: $\dfrac{8 x^5}{4 x^{1/2}} = 2 x^{5 - 1/2} = 2 x^{9/2}$ [1]. (b) At $x = 4$: $2 \cdot 4^{9/2} = 2 \cdot (\sqrt{4})^9 = 2 \cdot 2^9 = 2 \cdot 512 = 1024$ [1].
Unit Boss — Everything at once
Simplify $\dfrac{\left(2 a^{1/2} b^{-1}\right)^4 \cdot a^0 b^3}{8 a^{-1} b^{-2}}$ leaving your answer with positive indices and fractional indices where unavoidable. Then state which laws you used at each stage. Finally, comment: is the result a polynomial in $a$ and $b$?
Reveal solution
Bracket (power of a product): $\left(2 a^{1/2} b^{-1}\right)^4 = 2^4 \cdot a^{1/2 \times 4} \cdot b^{-1 \times 4} = 16 a^2 b^{-4}$. Zero index: $a^0 = 1$. Numerator product: $16 a^2 b^{-4} \cdot b^3 = 16 a^2 b^{-1}$. Quotient: $\dfrac{16 a^2 b^{-1}}{8 a^{-1} b^{-2}} = 2 \cdot a^{2 - (-1)} \cdot b^{-1 - (-2)} = 2 a^3 b$. The answer is $2 a^3 b$ — yes, this is a polynomial (no negative or fractional indices remain).
Product
$x^m x^n = x^{m+n}$
Quotient
$x^m / x^n = x^{m-n}$
Power
$(x^m)^n = x^{mn}$
Zero
$x^0 = 1$
Negative
$x^{-n} = 1/x^n$
Fractional
$x^{m/n} = \sqrt[n]{x^m}$
Sci. notation
$a \times 10^n$, $1 \le a < 10$
3 sig fig
Standard answer precision
Strategy
Powers $\to$ $\times$ $\to$ $\div$
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