2.1 Laws of indices for rational exponents — Practise

Work through these on your own. Before you evaluate a power, decide which to do first: the root or the power. Taking the root first keeps the numbers small.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Work out $16^{\frac{3}{4}}$ .
Work out $27^{-\frac{2}{3}}$ .
Work out $25^{\frac{3}{2}}$ .
Work out $32^{\frac{3}{5}}$ .
Work out $\left(\dfrac{16}{81}\right)^{\frac{3}{4}}$ .
Simplify $x^{\frac{1}{2}} \times x^{\frac{3}{2}}$ , where $x > 0$ .
Simplify $\dfrac{x^{\frac{5}{2}}}{x^{\frac{1}{2}}}$ , where $x > 0$ .
Simplify $\left(8x^6\right)^{\frac{2}{3}}$ , where $x > 0$ .
Solve $x^{\frac{2}{3}} = 9$ .
Solve $x^{\frac{3}{2}} = 27$ , where $x \ge 0$ .
Which is larger, $2^{\frac{1}{2}}$ or $2^{\frac{1}{3}}$ ? Give a reason.
A student writes $\sqrt{9 + 16} = \sqrt{9} + \sqrt{16}$ . Work out both sides and explain what the result shows.

Answers and concise solutions

1. The value

Answer: Take the fourth root first: $16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$ .

2. The value

Answer: The negative index gives a reciprocal: $27^{-\frac{2}{3}} = \dfrac{1}{27^{\frac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{27}\right)^2} = \dfrac{1}{3^2} = \dfrac{1}{9}$ .

3. The value

Answer: $25^{\frac{3}{2}} = \left(\sqrt{25}\right)^3 = 5^3 = 125$ .

4. The value

Answer: $32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$ . The fifth root of $32$ is $2$ because $2^5 = 32$ .

5. The value

Answer: The power applies to the numerator and the denominator: $\left(\dfrac{16}{81}\right)^{\frac{3}{4}} = \dfrac{16^{\frac{3}{4}}}{81^{\frac{3}{4}}} = \dfrac{8}{27}$ , since $16^{\frac{3}{4}} = 2^3 = 8$ and $81^{\frac{3}{4}} = 3^3 = 27$ . Ensure the cube is taken of the root, not of the base.

6. Simplified

Answer: Multiply by adding the indices: $x^{\frac{1}{2}} \times x^{\frac{3}{2}} = x^{\frac{1}{2} + \frac{3}{2}} = x^{2}$ . The condition $x > 0$ keeps every root real.

7. Simplified

Answer: Divide by subtracting the indices: $\dfrac{x^{\frac{5}{2}}}{x^{\frac{1}{2}}} = x^{\frac{5}{2} - \frac{1}{2}} = x^{2}$ .

8. Simplified

Answer: The power applies to each factor: $\left(8x^6\right)^{\frac{2}{3}} = 8^{\frac{2}{3}} \times \left(x^6\right)^{\frac{2}{3}} = 4x^{4}$ . The power of a power multiplies the indices: $6 \times \frac{2}{3} = 4$ .

9. The solutions

Answer: Raise both sides to the power $\frac{3}{2}$ : $x = 9^{\frac{3}{2}} = \left(\sqrt{9}\right)^3 = 27$ . The numerator $2$ is even, so a negative base gives the same value: $(-27)^{\frac{2}{3}} = \left(\sqrt[3]{-27}\right)^2 = 9$ . Both values solve the equation, so $x = 27$ or $x = -27$ .

10. The solution

Answer: Raise both sides to the power $\frac{2}{3}$ : $x = 27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 9$ . A negative base has no real value for $x^{\frac{3}{2}}$ , because the square root of a negative number is not real, so $x = 9$ is the only real solution.

11. The larger value, with a reason

Answer: $2^{\frac{1}{2}} = \sqrt{2} \approx 1.414$ and $2^{\frac{1}{3}} = \sqrt[3]{2} \approx 1.260$ . So $2^{\frac{1}{2}}$ is larger. For a base greater than $1$ , a larger index gives a larger value, and $\frac{1}{2} > \frac{1}{3}$ .

12. Both sides, and the conclusion

Answer: The left side is $\sqrt{25} = 5$ . The right side is $3 + 4 = 7$ . The sides are not equal, so a root cannot be split across a sum. The law $a^{\frac{1}{n}} b^{\frac{1}{n}} = (ab)^{\frac{1}{n}}$ applies to a product, not to a sum.