$$ \begin{array}{l} \left(2^{4} x^{4}\right) \ =\frac{3^{-3} y^{-6}}{2^{-3} x^{-3}} \ =\frac{2^{3} x^{3}}{3^{3} y^{6}} \ =\frac{8 x^{3}}{27 y^{6}} \end{array} $$ 26 To simplify expressions containing roots, we rewrite the roots as powers with rational exponents: EXAMPLE 3 Simplify: (a) $$ \frac{\sqrt{x^{3}} \times \sqrt[3]{x^{2}}}{\sqrt[6]{x}} $$ (b) $$ \sqrt[3]{8 x^{2} y^{6}} \times x^{\frac{1}{3}} $$ Solution (a) $$ \frac{x^{\frac{3}{2}} \times x^{\frac{2}{3}}}{x^{\frac{1}{6}}} $$ $$ \text { (b) } \quad\left(8 x^{2} y^{6}\right)^{\frac{1}{3}} \times x^{\frac{1}{3}} $$ $$ =x^{\frac{3}{2}+\frac{2}{3}-\frac{1}{6}} $$ $$ =x^{\frac{9+4-1}{6}} $$ $$ =x^{\frac{12}{6}}=x^{2} $$ Note: When applying roots to expressions, it is very important to distinguish between the behaviour of coefficients and exponents: $$ \sqrt{16 x^{16}}=4 x^{8} \quad $$ Take the square root of the coefficient, but divide the exponent by 2 . $$ \sqrt[3]{27 x^{27}}=3 x^{9} \quad $$ Take the cube root of the coefficient, but divide the exponent by 3 . EXERCISE 1 (a) Calculate the following, without the use of a calculator: (1) $$ 25^{\frac{1}{2}} $$ (2) $$ 8^{\frac{2}{3}} $$ (3) $$ 9^{-\frac{3}{2}} $$ (4) $$ 32^{-\frac{2}{5}} $$ (5) $$ \left(\frac{1}{8}\right)^{\frac{1}{3}} $$ (6) $$ \left(\frac{4}{9}\right)^{\frac{3}{2}} $$ (7) $$ \left(\frac{27}{8}\right)^{-\frac{2}{3}} $$ (8) $$ \left(\frac{16}{81}\right)^{-\frac{3}{4}} $$ (9) $$ (0,25)^{\frac{1}{2}} $$ (10) $$ \quad(0,125)^{\frac{2}{3}} $$ (11) $$ \quad(2,25)^{-\frac{1}{2}} $$ $$ (12)^{*} \quad(0, i)^{\frac{3}{2}} $$

Question

UpStudy AI · Accepted Answer

Sure, let's work through **EXERCISE 1(a)** step by step. We'll simplify each expression by rewriting roots as fractional exponents and applying exponent rules.

### **Exercise 1(a)**

**Calculate the following, without the use of a calculator:**

---

**1. $$ 25^{\frac{1}{2}} $$**

**Solution:**

The exponent $$\frac{1}{2}$$ represents the square root.

$$
25^{\frac{1}{2}} = \sqrt{25} = 5
$$

---

**2. $$ 8^{\frac{2}{3}} $$**

**Solution:**

First, interpret the exponent $$\frac{2}{3}$$ as the cube root raised to the power of 2.

$$
8^{\frac{2}{3}} = \left(8^{\frac{1}{3}}\right)^2 = (2)^2 = 4
$$

---

**3. $$ 9^{-\frac{3}{2}} $$**

**Solution:**

First, handle the negative exponent, then the fractional exponent.

$$
9^{-\frac{3}{2}} = \frac{1}{9^{\frac{3}{2}}} = \frac{1}{\left(\sqrt{9}\right)^3} = \frac{1}{3^3} = \frac{1}{27}
$$

---

**4. $$ 32^{-\frac{2}{5}} $$**

**Solution:**

Handle the negative exponent first, then the fractional exponent.

$$
32^{-\frac{2}{5}} = \frac{1}{32^{\frac{2}{5}}} = \frac{1}{\left(32^{\frac{1}{5}}\right)^2}
$$

Since $$32 = 2^5$$:

$$
32^{\frac{1}{5}} = 2 \Rightarrow \left(32^{\frac{1}{5}}\right)^2 = 2^2 = 4
$$

Thus:

$$
32^{-\frac{2}{5}} = \frac{1}{4}
$$

---

**5. $$ \left(\frac{1}{8}\right)^{\frac{1}{3}} $$**

**Solution:**

$$
\left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}} = \frac{1}{2} = \frac{1}{2}
$$

---

**6. $$ \left(\frac{4}{9}\right)^{\frac{3}{2}} $$**

**Solution:**

First, interpret the exponent $$\frac{3}{2}$$ as the square root raised to the power of 3.

$$
\left(\frac{4}{9}\right)^{\frac{3}{2}} = \left(\left(\frac{4}{9}\right)^{\frac{1}{2}}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}
$$

---

**7. $$ \left(\frac{27}{8}\right)^{-\frac{2}{3}} $$**

**Solution:**

Handle the negative exponent first.

$$
\left(\frac{27}{8}\right)^{-\frac{2}{3}} = \left(\frac{8}{27}\right)^{\frac{2}{3}}
$$

Now, apply the fractional exponent:

$$
\left(\frac{8}{27}\right)^{\frac{2}{3}} = \left(\frac{8^{\frac{1}{3}}}{27^{\frac{1}{3}}}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}
$$

---

**8. $$ \left(\frac{16}{81}\right)^{-\frac{3}{4}} $$**

**Solution:**

Handle the negative exponent first.

$$
\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left(\frac{81}{16}\right)^{\frac{3}{4}}
$$

Now, interpret the exponent $$\frac{3}{4}$$:

$$
\left(\frac{81}{16}\right)^{\frac{3}{4}} = \left(\left(\frac{81}{16}\right)^{\frac{1}{4}}\right)^3
$$

Since $$81 = 3^4$$ and $$16 = 2^4$$:

$$
\left(\frac{81}{16}\right)^{\frac{1}{4}} = \frac{3}{2} \Rightarrow \left(\frac{3}{2}\right)^3 = \frac{27}{8}
$$

Thus:

$$
\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \frac{27}{8}
$$

---

**9. $$ (0.25)^{\frac{1}{2}} $$**

**Solution:**

The exponent $$\frac{1}{2}$$ represents the square root.

$$
(0.25)^{\frac{1}{2}} = \sqrt{0.25} = 0.5
$$

---

**10. $$ (0.125)^{\frac{2}{3}} $$**

**Solution:**

First, interpret the exponent $$\frac{2}{3}$$ as the cube root raised to the power of 2.

$$
0.125 = \frac{1}{8} = 8^{-1}
$$

$$
(0.125)^{\frac{2}{3}} = \left(8^{-1}\right)^{\frac{2}{3}} = 8^{-\frac{2}{3}} = \left(8^{\frac{1}{3}}\right)^{-2} = (2)^{-2} = \frac{1}{4} = 0.25
$$

---

**11. $$ (2.25)^{-\frac{1}{2}} $$**

**Solution:**

Handle the negative exponent first.

$$
(2.25)^{-\frac{1}{2}} = \frac{1}{(2.25)^{\frac{1}{2}}} = \frac{1}{\sqrt{2.25}}
$$

Since $$2.25 = \frac{9}{4}$$:

$$
\sqrt{2.25} = \sqrt{\frac{9}{4}} = \frac{3}{2}
$$

Thus:

$$
(2.25)^{-\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}
$$

---

**12. $$ (0.1)^{\frac{3}{2}} $$ (*Note: Assuming the notation (0, i)^{\frac{3}{2}} was intended to be (0.1)^{\frac{3}{2}})*

**Solution:**

Interpret the exponent $$\frac{3}{2}$$ as the square root raised to the power of 3.

$$
(0.1)^{\frac{3}{2}} = \left(\sqrt{0.1}\right)^3
$$

First, find $$\sqrt{0.1}$$:

$$
\sqrt{0.1} = \sqrt{\frac{1}{10}} = \frac{\sqrt{10}}{10} \approx 0.3162
$$

Then, raise it to the power of 3:

$$
\left(\frac{\sqrt{10}}{10}\right)^3 = \frac{10^{\frac{3}{2}}}{10^3} = \frac{10^{1.5}}{1000} = \frac{31.6227766}{1000} \approx 0.0316
$$

**Alternatively**, keeping it exact:

$$
(0.1)^{\frac{3}{2}} = \left(\frac{1}{10}\right)^{\frac{3}{2}} = \frac{1^{\frac{3}{2}}}{10^{\frac{3}{2}}} = \frac{1}{10^{1.5}} = \frac{1}{10 \times \sqrt{10}} = \frac{\sqrt{10}}{100} \approx 0.0316
$$

---

**Summary of Answers:**

1. $$25^{\frac{1}{2}} = 5$$
2. $$8^{\frac{2}{3}} = 4$$
3. $$9^{-\frac{3}{2}} = \frac{1}{27}$$
4. $$32^{-\frac{2}{5}} = \frac{1}{4}$$
5. $$\left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$$
6. $$\left(\frac{4}{9}\right)^{\frac{3}{2}} = \frac{8}{27}$$
7. $$\left(\frac{27}{8}\right)^{-\frac{2}{3}} = \frac{4}{9}$$
8. $$\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \frac{27}{8}$$
9. $$(0.25)^{\frac{1}{2}} = 0.5$$
10. $$(0.125)^{\frac{2}{3}} = 0.25$$
11. $$(2.25)^{-\frac{1}{2}} = \frac{2}{3}$$
12. $$(0.1)^{\frac{3}{2}} \approx 0.0316$$

---

**Note on Exponents and Roots:**

- An exponent of $$\frac{1}{n}$$ represents the $$n$$th root.
- Negative exponents indicate reciprocals.
- Combine exponents using $$a^{m} \times a^{n} = a^{m+n}$$ and $$\left(a^{m}\right)^{n} = a^{m \times n}$$.

This exercise demonstrates how to manipulate and simplify expressions with rational exponents by applying these fundamental exponent rules.
Here are the simplified answers for each part of **EXERCISE 1(a)**: 1. $$25^{\frac{1}{2}} = 5$$ 2. $$8^{\frac{2}{3}} = 4$$ 3. $$9^{-\frac{3}{2}} = \frac{1}{27}$$ 4. $$32^{-\frac{2}{5}} = \frac{1}{4}$$ 5. $$\left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$$ 6. $$\left(\frac{4}{9}\right)^{\frac{3}{2}} = \frac{8}{27}$$ 7. $$\left(\frac{27}{8}\right)^{-\frac{2}{3}} = \frac{4}{9}$$ 8. $$\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \frac{27}{8}$$ 9. $$(0.25)^{\frac{1}{2}} = 0.5$$ 10. $$(0.125)^{\frac{2}{3}} = 0.25$$ 11. $$(2.25)^{-\frac{1}{2}} = \frac{2}{3}$$ 12. $$(0.1)^{\frac{3}{2}} \approx 0.0316$$ **Summary:** Each expression has been simplified by converting roots to fractional exponents and applying the rules of exponents to find the numerical values without using a calculator.

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Algebra Questions