Mathematics • Year 9 • Unit 1 • Lesson 20
Index Laws, Synthesis Build
Build fluency picking the right law from the full Unit 1 toolkit: product, quotient, power-of-a-power, power-of-a-product, zero, negative, and fractional. Plus scientific notation. Eight graduated problems take you from single-law warm-ups to four-law combinations.
1. I do, fully worked example
Read every line. This example uses five different index laws in a single expression, the kind of question the synthesis lesson is preparing you for.
Problem. Simplify $\dfrac{(2 a^3)^2 \cdot a^{-1}}{4 a^{1/2}}$, giving your answer with a positive index.
Step 1, Power-of-a-product on the bracket.
$(2 a^3)^2 = 2^2 \cdot (a^3)^2 = 4 a^{3 \times 2} = 4 a^6$
Reason: $(xy)^n = x^n y^n$, every factor inside the bracket gets the outer power. The $2$ becomes $2^2 = 4$ and the $a^3$ becomes $a^{3 \times 2} = a^6$ (power-of-a-power).
Step 2, Product rule on the numerator.
$4 a^6 \cdot a^{-1} = 4 a^{6 + (-1)} = 4 a^5$
Reason: $x^m \cdot x^n = x^{m+n}$. Add the indices, including the negative one.
Step 3, Quotient rule with the fractional index.
$\dfrac{4 a^5}{4 a^{1/2}} = \dfrac{4}{4} \cdot a^{5 - 1/2}$
Reason: $\dfrac{x^m}{x^n} = x^{m-n}$ works for any indices, including a fraction.
Step 4, Subtract the indices with a common denominator.
$5 - \dfrac{1}{2} = \dfrac{10}{2} - \dfrac{1}{2} = \dfrac{9}{2}$
Reason: write the whole number $5$ as $\dfrac{10}{2}$ so the subtraction works.
Step 5, Final form.
$1 \cdot a^{9/2} = a^{9/2}$
Reason: coefficient $\dfrac{4}{4} = 1$ disappears. Index is positive, so we're done, no negative-to-flip required.
Answer: $\mathbf{a^{9/2}}$ (equivalently $\sqrt{a^9}$ or $a^4 \sqrt{a}$).
2. We do, fill in the missing steps
Same structure as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. Simplify $\dfrac{6 a^5 b^{-2}}{2 a^2 b^{-5}}$, with positive indices.
Step 1, Divide the coefficients:
$\dfrac{6}{2} = \_\_\_\_$
Step 2, Quotient rule on $a$: subtract indices:
$a^{5 - \_\_} = a^{\_\_}$
Step 3, Quotient rule on $b$: watch the double negative:
$b^{-2 - (\_\_\_)} = b^{-2 + \_\_\_} = b^{\_\_\_}$
Step 4, Combine:
$\dfrac{6 a^5 b^{-2}}{2 a^2 b^{-5}} = \_\_\_\_\_ \cdot a^{\_\_} \cdot b^{\_\_}$
Step 5, Are all indices positive? __________ (yes / no)
3. You do, independent practice
Show your working under each problem. The first four are foundation (one or two laws). The middle two are standard (combining 2-3 laws). The last two are extension (4+ laws in one expression).
Foundation, single rule
3.1 Simplify $x^7 \cdot x^{-3}$ as a positive index. 1 mark
3.2 Evaluate $5^0 + 5^{-1}$. 1 mark
3.3 Simplify $\dfrac{x^8}{x^3}$. 1 mark
3.4 Expand $(3 x^2)^3$. 1 mark
Standard, two or three laws
3.5 Simplify $\dfrac{(3 x^2 y)^2}{9 x y}$. 2 marks
3.6 Evaluate $125^{2/3}$. 2 marks
Extension, four or more laws
3.7 Simplify $\dfrac{(2 x^2)^3 \cdot x^{-1}}{4 x^{1/2}}$, with positive indices. (Hint: power-of-a-product on the bracket, product rule on the numerator, quotient rule for the fraction, watch the fractional index.) 3 marks
3.8 Calculate $(2.5 \times 10^4) \times (4 \times 10^{-7})$, in standard scientific notation, then state the order of magnitude. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (faded $\dfrac{6 a^5 b^{-2}}{2 a^2 b^{-5}}$)
Step 1: $\dfrac{6}{2} = \mathbf{3}$.
Step 2: $a^{5 - \mathbf{2}} = a^{\mathbf{3}}$.
Step 3: $b^{-2 - (\mathbf{-5})} = b^{-2 + \mathbf{5}} = b^{\mathbf{3}}$.
Step 4: $\mathbf{3} \cdot a^{\mathbf{3}} \cdot b^{\mathbf{3}}$, i.e. $\mathbf{3 a^3 b^3}$.
Step 5: yes, all indices positive.
3.1, $x^7 \cdot x^{-3}$
Product rule: $x^{7 + (-3)} = \mathbf{x^4}$. (Positive index already.)
3.2, $5^0 + 5^{-1}$
$5^0 = 1$ (zero index). $5^{-1} = \dfrac{1}{5} = 0.2$ (negative index). Sum: $1 + 0.2 = \mathbf{1.2}$ (or $\dfrac{6}{5}$).
3.3, $\dfrac{x^8}{x^3}$
Quotient rule: $x^{8 - 3} = \mathbf{x^5}$.
3.4, $(3 x^2)^3$
Power-of-a-product: $3^3 \cdot (x^2)^3 = 27 \cdot x^{2 \times 3} = \mathbf{27 x^6}$.
3.5, $\dfrac{(3 x^2 y)^2}{9 x y}$
Numerator: $(3 x^2 y)^2 = 9 x^4 y^2$ (power-of-a-product). Now divide: $\dfrac{9 x^4 y^2}{9 x y} = 1 \cdot x^{4 - 1} \cdot y^{2 - 1} = \mathbf{x^3 y}$.
3.6, $125^{2/3}$
Root first: $\sqrt[3]{125} = 5$ (since $5^3 = 125$). Then $5^2 = \mathbf{25}$.
3.7, $\dfrac{(2 x^2)^3 \cdot x^{-1}}{4 x^{1/2}}$
Step 1, power-of-a-product: $(2 x^2)^3 = 8 x^6$.
Step 2, product on numerator: $8 x^6 \cdot x^{-1} = 8 x^{6 - 1} = 8 x^5$.
Step 3, quotient: $\dfrac{8 x^5}{4 x^{1/2}} = 2 \cdot x^{5 - 1/2}$.
Step 4, common denominator: $5 - \dfrac{1}{2} = \dfrac{10}{2} - \dfrac{1}{2} = \dfrac{9}{2}$.
Answer: $\mathbf{2 x^{9/2}}$ (positive index).
3.8, $(2.5 \times 10^4) \times (4 \times 10^{-7})$
Coefficients: $2.5 \times 4 = 10$. Indices: $10^{4 + (-7)} = 10^{-3}$. Current: $10 \times 10^{-3}$. Re-normalise: $\mathbf{1 \times 10^{-2}}$ (or $0.01$). Order of magnitude $\approx \mathbf{10^{-2}}$.