1. Solve the following exponential equations: i. Solve $3^{3-x} = 27^{x-1}$. - Express 27 as a power of 3: $27 = 3^3$. - Rewrite the equation: $3^{3-x} = (3^3)^{x-1} = 3^{3(x-1)}$. - Equate exponents: $3 - x = 3(x - 1)$. - Simplify: $3 - x = 3x - 3$. - Rearrange: $3 + 3 = 3x + x$ gives $6 = 4x$. - Solve: $x = \frac{6}{4} = 1.5$. ii. Solve $4^{x+1} = 0.125$. - Express 4 and 0.125 as powers of 2: $4 = 2^2$, $0.125 = \frac{1}{8} = 2^{-3}$. - Rewrite the equation: $(2^2)^{x+1} = 2^{-3}$ or $2^{2(x+1)} = 2^{-3}$. - Equate exponents: $2x + 2 = -3$. - Solve: $2x = -5$ so $x = -\frac{5}{2} = -2.5$. iii. Solve $5^{4x} = 27^{x+3}$. - Express 27 as powers of 3: $27 = 3^3$. - So equation is $5^{4x} = (3^3)^{x+3} = 3^{3(x+3)}$. - Take logarithm base 10 or natural log on both sides: $4x \log 5 = 3(x + 3) \log 3$. - Expand: $4x \log 5 = 3x \log 3 + 9 \log 3$. - Rearrange: $4x \log 5 - 3x \log 3 = 9 \log 3$. - Factor x: $x(4 \log 5 - 3 \log 3) = 9 \log 3$. - Solve for x: $$x = \frac{9 \log 3}{4 \log 5 - 3 \log 3}$$. 2. Solve the following logarithmic equations: i. Solve $\log_2(3x + 1) - \log_2(2x - 7) = 3$. - Use log subtraction property: $\log_2 \left( \frac{3x + 1}{2x - 7} \right) = 3$. - Rewrite in exponential form: $\frac{3x + 1}{2x - 7} = 2^3 = 8$. - Solve for x: $$3x + 1 = 8(2x - 7) = 16x - 56$$ $$3x + 1 = 16x - 56$$ $$1 + 56 = 16x - 3x$$ $$57 = 13x$$ $$x = \frac{57}{13}$$. - Check domain: $2x - 7 > 0 \Rightarrow x > 3.5$, and $3x + 1 > 0$. Since $\frac{57}{13} \approx 4.38 > 3.5$, solution valid. ii. Solve $2 + \log_3 (x-1) = \log_3 x$. - Rewrite 2 as $\log_3 9$: since $3^2=9$. - Equation becomes: $\log_3 9 + \log_3 (x-1) = \log_3 x$. - Use log addition property: $\log_3 [9(x-1)] = \log_3 x$. - Equate arguments: $9(x-1) = x$. - Expand: $9x - 9 = x$. - Rearrange: $9x - x = 9$. - Simplify: $8x = 9$. - Solve: $x = \frac{9}{8} = 1.125$. - Check domain: $x-1 > 0 \Rightarrow x>1$ and $x>0$. Solution valid. 3. Express $\log_5 9.8$ in terms of $m = \log_5 2$ and $p = \log_5 7$. - Write 9.8 as $9.8 = \frac{98}{10} = \frac{7 \times 14}{10}$. Better to factor directly: - $9.8 = 2 \times 4.9 = 2 \times (7 \times 0.7)$ but 0.7 is messy. - Instead, express $9.8 = \frac{49}{5} = \frac{7^2}{5}$. - So, $\log_5 9.8 = \log_5 \left( \frac{7^2}{5} \right) = \log_5 7^2 - \log_5 5 = 2 \log_5 7 - 1$ (since $\log_5 5 =1$). - Recall $\log_5 7 = p$, so final expression: $$\log_5 9.8 = 2p - 1$$. 4. Simplify the following radicals: i. Simplify $\sqrt{243} - \sqrt{12} + 2\sqrt{75}$. - Break down each radical: $$\sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$$ $$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$$ $$2\sqrt{75} = 2\sqrt{25 \times 3} = 2 \times 5 \sqrt{3} = 10\sqrt{3}$$ - Combine: $$9\sqrt{3} - 2\sqrt{3} + 10\sqrt{3} = (9 - 2 + 10)\sqrt{3} = 17\sqrt{3}$$. ii. Simplify $\sqrt{50} + \sqrt{8} + \sqrt{0.32}$. - Break down each: $$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$ $$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$$ $$\sqrt{0.32} = \sqrt{\frac{32}{100}} = \frac{\sqrt{32}}{10} = \frac{\sqrt{16 \times 2}}{10} = \frac{4\sqrt{2}}{10} = 0.4\sqrt{2}$$ - Combine: $$5\sqrt{2} + 2\sqrt{2} + 0.4\sqrt{2} = (5 + 2 + 0.4)\sqrt{2} = 7.4\sqrt{2}$$. Final answers: - a i: $x = 1.5$ - a ii: $x = -2.5$ - a iii: $x = \frac{9 \log 3}{4 \log 5 - 3 \log 3}$ - b i: $x = \frac{57}{13}$ - b ii: $x = \frac{9}{8}$ - c: $\log_5 9.8 = 2p - 1$ - d i: $17 \sqrt{3}$ - d ii: $7.4 \sqrt{2}$