Let's solve each part **step by step**: --- ## (a) Solve as a single number raised to a power ### (i) \( 3^8 \times 3^{-5} \times 3^2 \) **Step 1:** Use the law of indices: \( a^m \times a^n = a^{m+n} \): \[ 3^8 \times 3^{-5} \times 3^2 = 3^{8 + (-5) + 2} \] **Step 2:** Simplify the exponents: \[ = 3^{8 - 5 + 2} = 3^{3 + 2} = 3^5 \] **Final Answer:** \[ \boxed{3^5} \] --- ### (ii) \( \frac{5^5}{5^2} \times 5^{-3} \times 5^3 \) **Step 1:** Use the laws of indices: \( \frac{a^m}{a^n} = a^{m-n} \) and \( a^m \times a^n = a^{m+n} \): \[ \frac{5^5}{5^2} = 5^{5-2} = 5^3 \] So the expression becomes: \[ 5^3 \times 5^{-3} \times 5^3 \] **Step 2:** Combine the exponents: \[ = 5^{3 + (-3) + 3} = 5^{0 + 3} = 5^3 \] **Final Answer:** \[ \boxed{5^3} \] --- ## (b) Simplify the following expressions ### (i) \( 12\sqrt{3} - 5\sqrt{3} + \sqrt{3} \) **Step 1:** Combine like terms (all are multiples of \( \sqrt{3} \)): \[ (12 - 5 + 1)\sqrt{3} = (7 + 1)\sqrt{3} = 8\sqrt{3} \] **Final Answer:** \[ \boxed{8\sqrt{3}} \] --- ### (ii) \( (2 + \sqrt{2}) \times (2 - \sqrt{2}) \) **Step 1:** Use the difference of squares formula: \( (a+b)(a-b) = a^2 - b^2 \): \[ (2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \] **Final Answer:** \[ \boxed{2} \] --- ## (c) Identify the polynomials, its terms and its coefficients in \( 2y^3 + 4y + 3 \) - **Polynomial:** \( 2y^3 + 4y + 3 \) - **Terms:** \( 2y^3, 4y, 3 \) - **Coefficients:** - Coefficient of \( y^3 \): **2** - Coefficient of \( y \): **4** - Coefficient of constant term (3): **3** **Final Answer:** - **Polynomial:** \( 2y^3 + 4y + 3 \) - **Terms:** \( 2y^3, 4y, 3 \) - **Coefficients:** 2, 4, 3 --- ## (d) Find the value of ### (i) \( \log_2 2 + \log_6 3 \) - \( \log_2 2 = 1 \) because \( 2^1 = 2 \) - \( \log_6 3 \) means the power to which 6 must be raised to get 3. Let’s express 3 in terms of 6: - \( 6^x = 3 \implies x = \log_6 3 \) - \( 3 = 6^{1/2} \) is not true, but \( 3 = 6^x \implies x = \log_6 3 \) - Alternatively, use change of base: \[ \log_6 3 = \frac{\log 3}{\log 6} \] but let's keep it as is for now. - So, the answer is \( 1 + \log_6 3 \). **If you want a decimal approximation:** \[ \log_6 3 = \frac{\log_{10} 3}{\log_{10} 6} \approx \frac{0.4771}{0.7782} \approx 0.613 \] So, \[ 1 + 0.613 = 1.613 \] **Final Answer:** \[ \boxed{1 + \log_6 3} \text{ or approximately } \boxed{1.613} \] --- ### (ii) \( \log_3 63 - \log_3 7 \) **Step 1:** Use log subtraction rule: \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \): \[ \log_3 63 - \log_3 7 = \log_3 \left( \frac{63}{7} \right) = \log_3 9 \] **Step 2:** \( 9 = 3^2 \implies \log_3 9 = 2 \) **Final Answer:** \[ \boxed{2} \] --- # **Summary Table of Answers** | Part | Answer | |:----:|:---------------------------| | (a)(i) | \( 3^5 \) | | (a)(ii) | \( 5^3 \) | | (b)(i) | \( 8\sqrt{3} \) | | (b)(ii) | \( 2 \) | | (c) | Polynomial: \( 2y^3 + 4y + 3 \), Terms: \( 2y^3, 4y, 3 \), Coefficients: 2, 4, 3 | | (d)(i) | \( 1 + \log_6 3 \) or \( 1.613 \) | | (d)(ii) | \( 2 \) |