Today's lesson
Learning intentions
- Convert between radical form ($\sqrt[m]{x}$) and rational-exponent form ($x^{1/m}$)
- Evaluate expressions with rational exponents like $27^{1/3}$, $16^{5/2}$, $64^{-2/3}$
- Handle negative bases (odd roots only!) and negative exponents
- Simplify mixed expressions using all index laws together
Work through every worked example below carefully — answers are shown so you can self-check your method. Then attempt the quiz and the Cambridge questions.
Part 1 — The first key rule
$x^{\frac{1}{m}} = \sqrt[m]{x}$
📺 Why? The index-law derivation, and an example: $27^{1/3}=3$.
A power of $\tfrac{1}{m}$ is the same as taking an $m$-th root. So $x^{1/2}=\sqrt{x}$, $x^{1/3}=\sqrt[3]{x}$, and so on.
Worked examples — Rewrite in index form
$\sqrt[2]{x}$
An index of $\tfrac{1}{m}$ where $m=2$.
$= x^{\frac{1}{2}}$
$\sqrt[4]{x}$
An index of $\tfrac{1}{m}$ where $m=4$.
$= x^{\frac{1}{4}}$
$\sqrt[7]{x}$
An index of $\tfrac{1}{m}$ where $m=7$.
$= x^{\frac{1}{7}}$
Worked examples — Evaluate
$9^{\frac{1}{2}}$
$= \sqrt{9}$
$= 3$
$27^{\frac{1}{3}}$
$= \sqrt[3]{27}$
What cubed gives 27? $3^3=27$
$= 3$
$32^{\frac{1}{5}}$
$= \sqrt[5]{32}$
What to the 5th gives 32? $2^5=32$
$= 2$
Part 2 — Extending to $x^{\frac{m}{n}}$
$x^{\frac{m}{n}} \;=\; \left(x^{\frac{1}{n}}\right)^{m} \;=\; \left(\sqrt[n]{x}\right)^{m} \;=\; \sqrt[n]{x^{m}}$
📺 Two paths for $64^{2/3}$ — taking the root first keeps the numbers small.
Pro tip: When evaluating, take the root first, then raise to the power. The numbers are much smaller and easier to handle. e.g. for $64^{2/3}$ do $(\sqrt[3]{64})^2 = 4^2 = 16$ rather than $\sqrt[3]{64^2} = \sqrt[3]{4096}$.
Worked examples — Rewrite
$\sqrt[5]{x^2}$
Outside index $=m=2$, root $=n=5$.
$= x^{\frac{2}{5}}$
$\left(\sqrt[3]{x}\right)^{2}$
Inside: $x^{1/3}$, then raised to 2.
$=\left(x^{\frac{1}{3}}\right)^{2}$
$= x^{\frac{2}{3}}$
$\sqrt[3]{x^{7}}$
$m=7$, $n=3$.
$= x^{\frac{7}{3}}$
Worked examples — Evaluate (root first, then power)
$16^{\frac{5}{2}}$
$=\left(16^{\frac{1}{2}}\right)^{5}$
$=\left(\sqrt{16}\right)^{5} = 4^{5}$
$= 1024$
$49^{\frac{3}{2}}$
$=\left(\sqrt{49}\right)^{3} = 7^{3}$
$= 343$
$64^{\frac{2}{3}}$
$=\left(\sqrt[3]{64}\right)^{2} = 4^{2}$
$= 16$
Part 3 — Negative bases & negative exponents
⚠️ Negative bases: You can only take an odd root of a negative number in the reals. $\sqrt[3]{-8} = -2$ is fine; $\sqrt{-4}$ is undefined in $\mathbb{R}$.
Negative exponents: $a^{-n} = \dfrac{1}{a^{n}}$ — the negative sign flips it to the denominator.
$(-64)^{\frac{1}{3}}$
$=\sqrt[3]{-64}$ (odd root, so OK)
What cubed gives $-64$? $(-4)^3 = -64$
$= -4$
$(-27)^{\frac{2}{3}}$
$=\left(\sqrt[3]{-27}\right)^{2}$
$=(-3)^{2}$
$= 9$
$9^{-\frac{1}{2}}$
$=\dfrac{1}{9^{\frac{1}{2}}}=\dfrac{1}{\sqrt{9}}$
$= \dfrac{1}{3}$
$64^{-\frac{2}{3}}$
$=\dfrac{1}{64^{\frac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{64}\right)^{2}}$
$=\dfrac{1}{4^{2}}$
$= \dfrac{1}{16}$
$125^{-\frac{4}{3}}$
$=\dfrac{1}{125^{\frac{4}{3}}} = \dfrac{1}{\left(\sqrt[3]{125}\right)^{4}}$
$=\dfrac{1}{5^{4}}$
$= \dfrac{1}{625}$
$\left(\dfrac{1}{36}\right)^{-\frac{3}{2}}$
Negative exponent flips the fraction: $=36^{\frac{3}{2}}$
$=\left(\sqrt{36}\right)^{3} = 6^{3}$
$= 216$
Part 4 — Mixed simplification (challenge)
Express your answer in positive-index form.
1. $\dfrac{3^{\frac{1}{4}} \times \sqrt{6}}{16^{\frac{3}{4}}} \div \dfrac{1}{\sqrt[4]{2}}$
Rewrite everything in prime-power form.
$\sqrt{6} = (2\cdot 3)^{\frac{1}{2}} = 2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}}$
$16^{\frac{3}{4}} = (2^{4})^{\frac{3}{4}} = 2^{3}$
$\sqrt[4]{2} = 2^{\frac{1}{4}}$ and $\div \tfrac{1}{2^{1/4}} = \times 2^{\frac{1}{4}}$
Combine: $\dfrac{3^{\frac{1}{4}} \cdot 2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}}}{2^{3}} \times 2^{\frac{1}{4}}$
Powers of 3: $3^{\frac{1}{4}+\frac{1}{2}} = 3^{\frac{3}{4}}$
Powers of 2: $2^{\frac{1}{2}+\frac{1}{4}-3} = 2^{\frac{3}{4}-3} = 2^{-\frac{9}{4}}$
So far: $3^{\frac{3}{4}} \cdot 2^{-\frac{9}{4}} = \dfrac{3^{\frac{3}{4}}}{2^{\frac{9}{4}}}$
Tidier: $\dfrac{3^{\frac{3}{4}} \cdot 2^{\frac{3}{4}}}{2^{3}} = \dfrac{(2\cdot 3)^{\frac{3}{4}}}{2^{3}} = \dfrac{6^{\frac{3}{4}}}{8}$
$=\dfrac{6^{\frac{3}{4}}}{8}$
2. $\left(x^{-2}y\right)^{\frac{1}{2}} \times \left(\dfrac{x}{y^{-3}}\right)^{4}$
First bracket: $\left(x^{-2}y\right)^{\frac{1}{2}} = x^{-1}y^{\frac{1}{2}}$
Inside second bracket: $\dfrac{x}{y^{-3}} = x \cdot y^{3}$
Second bracket: $(xy^{3})^{4} = x^{4}y^{12}$
Multiply: $x^{-1}y^{\frac{1}{2}} \cdot x^{4}y^{12} = x^{-1+4} \cdot y^{\frac{1}{2}+12}$
$= x^{3}y^{\frac{25}{2}}$
3. $(x+4)^{2}\sqrt[3]{x+4}$
Rewrite the cube root: $\sqrt[3]{x+4} = (x+4)^{\frac{1}{3}}$
$= (x+4)^{2} \cdot (x+4)^{\frac{1}{3}}$
Same base — add the indices: $2 + \tfrac{1}{3} = \tfrac{7}{3}$
$= (x+4)^{\frac{7}{3}}$
4. $\sqrt{2x+1} - \dfrac{1}{\sqrt{2x+1}}$
Common denominator $\sqrt{2x+1}$:
$= \dfrac{(\sqrt{2x+1})^{2}}{\sqrt{2x+1}} - \dfrac{1}{\sqrt{2x+1}}$
$= \dfrac{(2x+1) - 1}{\sqrt{2x+1}}$
$= \dfrac{2x}{\sqrt{2x+1}} \;\;\text{or}\;\; 2x(2x+1)^{-\frac{1}{2}}$
Part 5 — Quick quiz (mark yourself)
Pick the correct answer for each, then click Mark.
Q1. Rewrite $\sqrt[5]{x^{3}}$ in index form.
Q2. Evaluate $81^{\frac{3}{4}}$.
Q3. Evaluate $8^{-\frac{2}{3}}$.
Q4. Evaluate $(-32)^{\frac{1}{5}}$.
Q5. Evaluate $\left(\dfrac{1}{25}\right)^{-\frac{1}{2}}$.
Q6. Simplify $x^{\frac{2}{3}} \cdot x^{\frac{1}{6}}$ (positive-index form).
Q7. Simplify $(x+1)^{3}\sqrt{x+1}$.
Working program — Cambridge textbook
Open Cambridge Chapter 13 and complete these in your exercise book:
| Exercise | Set work |
|---|---|
| 13B — Rational Exponents | Q1*, Q2*, Q3* |
* = full question, all parts. If you finish, start 13C (Exponential Graphs).
Exit ticket — write in your book
Before you pack up, in your exercise book write:
- The general rule for $x^{\frac{m}{n}}$ (in radical form).
- Why is $(-25)^{\frac{1}{2}}$ undefined but $(-27)^{\frac{1}{3}}$ is fine?
- Evaluate (no calculator): $\;\;100^{-\frac{3}{2}}$