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
Print or save as PDF, or build a custom worksheet from any module's questions.
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 $\to$ $\times$ $\to$ $\div$
Your Badges
0 of 6Mark lesson as complete
Tick when you've finished Learn, Practice and the Stretch. Earns +95 XP and +30 coins, bonus for finishing the unit!