Fractional Indices (Extension)
Read $x^{1/2}$ as $\sqrt{x}$, $x^{1/3}$ as $\sqrt[3]{x}$, and unlock $x^{m/n} = \sqrt[n]{x^m}$. Connect surds to index laws and simplify the lot.
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Use the index laws you already know to predict $25^{1/2}$ and $8^{1/3}$. Hint: try $25^{1/2} \times 25^{1/2}$ first — what does the product rule say about the index?
A fractional index is a root in disguise. The bottom of the fraction tells you which root; the top tells you which power to take. All the index laws still apply.
$x^{1/n} = \sqrt[n]{x}$ — the denominator is the root. $x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$ — the numerator is the power. Same product, quotient and power rules as before.
Know
- $x^{1/2} = \sqrt{x}$ and $x^{1/3} = \sqrt[3]{x}$
- $x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$
- Index laws extend to fractional indices
Understand
- Why $x^{1/n} \times x^{1/n} \times \ldots = x$ forces $x^{1/n}$ to be the $n$th root
- Why “root first, then power” keeps the numbers small
- How surds and fractional indices are the same idea, two notations
Can Do
- Evaluate $16^{1/2}$, $27^{1/3}$, $8^{2/3}$
- Convert $\sqrt[3]{x^5}$ to index form and back
- Simplify $x^{1/2} \times x^{1/3}$ to $x^{5/6}$
Wrong: “$16^{1/2} = 8$” — halving instead of taking a root.
Right: $16^{1/2} = \sqrt{16} = 4$. The index $1/2$ is the square root, not “divide by $2$.”
Wrong: “$8^{2/3} = \sqrt[3]{8} \times 2 = 4$” — multiplying instead of cubing.
Right: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. The $2$ is a power, applied after the root.
For nice numbers, take the root first, then raise to the power. The numbers stay small and tidy.
$8^{2/3}$: cube root of $8$ is $2$; square it $\to 4$. $32^{3/5}$: fifth root of $32$ is $2$; cube it $\to 8$. $64^{2/3}$: cube root of $64$ is $4$; square it $\to 16$. Note: $8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$ — same answer, bigger intermediate.
Fractional indices behave like any other index: add to multiply, subtract to divide, multiply to raise to a power. Find a common denominator when adding fractions.
$x^{1/2} \times x^{1/3} = x^{1/2 + 1/3} = x^{3/6 + 2/6} = x^{5/6}$. $\dfrac{x^{3/4}}{x^{1/4}} = x^{3/4 - 1/4} = x^{2/4} = x^{1/2} = \sqrt{x}$. $(x^{2/3})^6 = x^{2/3 \times 6} = x^4$.
Watch Me Solve It · 3 examples
- 1Identify root and powerDenominator $3 \to$ cube root; numerator $2 \to$ square
- 2Take the cube root first$\sqrt[3]{27} = 3$Keeps the number small.
- 3Square the result$3^2 = 9$
- 1Convert the surd$\sqrt[4]{x^5} = x^{5/4}$Power on top, root on bottom.
- 2Apply product rule$x^{5/4} \times x^{1/4} = x^{5/4 + 1/4}$
- 3Add the fractions$x^{6/4} = x^{3/2}$
- 1Product rule on the numerator$x^{2/3} \cdot x^3 = x^{2/3 + 3} = x^{2/3 + 9/3} = x^{11/3}$
- 2Quotient rule with the denominator$\dfrac{x^{11/3}}{x^{1/2}} = x^{11/3 - 1/2}$
- 3Common denominator and subtract$\dfrac{11}{3} - \dfrac{1}{2} = \dfrac{22}{6} - \dfrac{3}{6} = \dfrac{19}{6}$. So $x^{19/6}$.
Common Pitfalls
Definitions
- $x^{1/n} = \sqrt[n]{x}$
- $x^{m/n} = \sqrt[n]{x^m}$
Key examples
- $16^{1/2} = 4$
- $27^{1/3} = 3$
- $8^{2/3} = 4$
Index laws still apply
- $x^{1/2} \cdot x^{1/3} = x^{5/6}$
- $(x^{2/3})^6 = x^4$
Strategy
- Root first $\to$ then power
- Common denominator to add indices
How are you completing this lesson?
Brain Trainer · 4 problems
Evaluate, convert and simplify. Use the “root first” strategy whenever you can.
1 Evaluate $36^{1/2}$.
Bottom is $2$, so square root of $36$.$6$2 Evaluate $125^{2/3}$.
Cube root of $125$ is $5$; square it: $5^2 = 25$.$25$3 Write $\sqrt[3]{x^5}$ in index form.
Power on top, root on bottom.$x^{5/3}$4 Simplify $x^{3/4} \cdot x^{1/2}$.
$\dfrac{3}{4} + \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{2}{4} = \dfrac{5}{4}$.$x^{5/4}$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Evaluate without a calculator: (a) $81^{1/2}$, (b) $32^{1/5}$, (c) $16^{3/4}$.
Q7. Convert and simplify, leaving the answer in index form: (a) $\sqrt{x} \cdot \sqrt[3]{x}$, (b) $\dfrac{\sqrt[4]{x^7}}{\sqrt{x}}$, (c) $\left(\sqrt[3]{x^2}\right)^6$.
Q8. Use index laws to prove $x^{1/2} \cdot x^{1/2} = x$, then explain in words why this forces $x^{1/2}$ to be defined as $\sqrt{x}$. Hence evaluate $\sqrt{144 x^4}$ in index form.
Quick Check
1. C — $7$.
2. A — $16$.
3. D — $x^{2/5}$.
4. B — $x^{5/6}$.
5. A — $x^4$.
Show Your Working Model Answers
Q6 (3 marks): (a) $\sqrt{81} = 9$ [1]; (b) $\sqrt[5]{32} = 2$ [1]; (c) $\sqrt[4]{16} = 2$, then $2^3 = 8$ [1].
Q7 (3 marks): (a) $x^{1/2} \cdot x^{1/3} = x^{3/6 + 2/6} = x^{5/6}$ [1]; (b) $x^{7/4} / x^{1/2} = x^{7/4 - 2/4} = x^{5/4}$ [1]; (c) $(x^{2/3})^6 = x^{12/3} = x^4$ [1].
Q8 (3 marks): Product rule: $x^{1/2} \cdot x^{1/2} = x^{1/2 + 1/2} = x^1 = x$ [1]. So $x^{1/2}$ is a number that, when multiplied by itself, gives $x$ — that's the definition of $\sqrt{x}$ [1]. Hence $\sqrt{144 x^4} = (144 x^4)^{1/2} = 144^{1/2} \cdot x^{4 \times 1/2} = 12 x^2$ [1].
Negative-Fractional Combo
Use index laws to simplify $\left(\dfrac{x^{1/2} \cdot x^{-1/3}}{x^{2/3}}\right)^{6}$, leaving your answer with a positive index. Then evaluate when $x = 64$.
Reveal solution
Inside index: $\dfrac{1}{2} - \dfrac{1}{3} - \dfrac{2}{3} = \dfrac{3}{6} - \dfrac{2}{6} - \dfrac{4}{6} = -\dfrac{3}{6} = -\dfrac{1}{2}$. So inside $= x^{-1/2}$. Raise to the $6$th: $x^{-3} = \dfrac{1}{x^3}$. At $x = 64$: $\dfrac{1}{64^3} = \dfrac{1}{262{,}144}$.
$x^{1/2}$
$= \sqrt{x}$
$x^{1/3}$
$= \sqrt[3]{x}$
$x^{m/n}$
$= \sqrt[n]{x^m}$
$8^{2/3}$
$= (\sqrt[3]{8})^2 = 4$
Root first
then take the power
Index laws hold
$x^{1/2} \cdot x^{1/3} = x^{5/6}$
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