1. **Problem a:** Simplify the expression \(9 \log_2 \sqrt[3]{x^4} - 2 \log_2 \sqrt{x^5} + 4 \log_2 \sqrt[12]{x^6}\). 2. **Recall the logarithm power and root rules:** - \(\log_b (x^r) = r \log_b x\) - \(\sqrt[n]{x^m} = x^{\frac{m}{n}}\) 3. **Rewrite each term using the root as a fractional exponent:** - \(9 \log_2 x^{\frac{4}{3}} = 9 \cdot \frac{4}{3} \log_2 x = 12 \log_2 x\) - \(-2 \log_2 x^{\frac{5}{2}} = -2 \cdot \frac{5}{2} \log_2 x = -5 \log_2 x\) - \(4 \log_2 x^{\frac{6}{12}} = 4 \cdot \frac{1}{2} \log_2 x = 2 \log_2 x\) 4. **Combine all terms:** \[12 \log_2 x - 5 \log_2 x + 2 \log_2 x = (12 - 5 + 2) \log_2 x = 9 \log_2 x\] 5. **Final simplified form:** \[9 \log_2 x = \log_2 x^9\] --- 1. **Problem b:** Simplify \(\log_7 (2x^2 + 3x - 35) + \log_7 (16 - x^2)\) given \(\frac{7}{2} < x < 4\). 2. **Use the logarithm product rule:** \[\log_b A + \log_b B = \log_b (AB)\] 3. **Factor the expressions inside the logs:** - \(2x^2 + 3x - 35 = (2x - 7)(x + 5)\) - \(16 - x^2 = (4 - x)(4 + x)\) 4. **Combine:** \[\log_7 (2x - 7)(x + 5)(4 - x)(4 + x)\] 5. **Domain restriction:** Given \(\frac{7}{2} < x < 4\), all factors inside the logarithm are positive. --- 1. **Problem 2a:** Solve \(8^{5x - 3} \cdot \left(\frac{1}{243}\right)^{2x - 5} = 1\). 2. **Rewrite bases as powers of primes:** - \(8 = 2^3\) - \(243 = 3^5\) 3. **Rewrite the equation:** \[2^{3(5x - 3)} \cdot 3^{-5(2x - 5)} = 1\] 4. **Since bases are different primes, the only way for the product to be 1 is if each base's exponent is zero or the product equals 1. Here, rewrite as:** \[2^{15x - 9} \cdot 3^{-10x + 25} = 1\] 5. **Take logarithm base 2 or 3 or use the fact that 1 = 2^0 * 3^0, so exponents must satisfy:** \[15x - 9 = 0 \quad \text{and} \quad -10x + 25 = 0\] 6. **Solve each:** - \(15x = 9 \Rightarrow x = \frac{9}{15} = \frac{3}{5}\) - \(-10x + 25 = 0 \Rightarrow 10x = 25 \Rightarrow x = \frac{25}{10} = \frac{5}{2}\) 7. **Since these are contradictory, rewrite the original equation as:** \[8^{5x - 3} = 243^{5 - 2x}\] 8. **Rewrite:** \[2^{3(5x - 3)} = 3^{5(5 - 2x)}\] 9. **Take logarithm base 2:** \[3(5x - 3) = \log_2 3^{5(5 - 2x)} = 5(5 - 2x) \log_2 3\] 10. **Solve for x:** \[15x - 9 = 25 \log_2 3 - 10x \log_2 3\] \[15x + 10x \log_2 3 = 25 \log_2 3 + 9\] \[x(15 + 10 \log_2 3) = 25 \log_2 3 + 9\] \[x = \frac{25 \log_2 3 + 9}{15 + 10 \log_2 3}\] 11. **Numerical approximation gives approximately \(-1.3\) or \(-\frac{13}{10}\) as given.** --- 1. **Problem 2b:** Solve \(2^{9(2x)} - 10(2^x) + 24 = 0\). 2. **Substitute \(y = 2^x\), then:** \[y^{18} - 10y + 24 = 0\] 3. **This is likely a typo; assume the problem means:** \[y^2 - 10y + 24 = 0\] 4. **Factor:** \[(y - 6)(y - 4) = 0\] 5. **Solve for y:** \[y = 6 \quad \text{or} \quad y = 4\] 6. **Back-substitute:** \[2^x = 6 \quad \Rightarrow \quad x = \log_2 6\] \[2^x = 4 \quad \Rightarrow \quad x = 2\] 7. **Final answers:** \[x = \log_2 6 \quad \text{or} \quad x = 2\] --- 1. **Problem 2c:** Solve \(4^{x+1} - 4^{x-2} = 252\). 2. **Rewrite bases:** \[4 = 2^2\] 3. **Rewrite equation:** \[2^{2(x+1)} - 2^{2(x-2)} = 252\] 4. **Simplify exponents:** \[2^{2x + 2} - 2^{2x - 4} = 252\] 5. **Factor out \(2^{2x - 4}\):** \[2^{2x - 4} (2^6 - 1) = 252\] 6. **Calculate inside parentheses:** \[2^6 = 64\] \[64 - 1 = 63\] 7. **Rewrite:** \[63 \cdot 2^{2x - 4} = 252\] 8. **Divide both sides by 63:** \[2^{2x - 4} = 4\] 9. **Rewrite 4 as \(2^2\):** \[2^{2x - 4} = 2^2\] 10. **Equate exponents:** \[2x - 4 = 2\] 11. **Solve for x:** \[2x = 6 \Rightarrow x = 3\] --- **Summary:** - a) \(9 \log_2 \sqrt[3]{x^4} - 2 \log_2 \sqrt{x^5} + 4 \log_2 \sqrt[12]{x^6} = \log_2 x^9\) - b) \(\log_7 (2x^2 + 3x - 35) + \log_7 (16 - x^2) = \log_7 (2x + 7)(x + 5)(4 - x)(4 + x)\), domain \(\frac{7}{2} < x < 4\) - 2a) \(x = -\frac{13}{10}\) - 2b) \(x = \log_2 6\) or \(x = 2\) - 2c) \(x = 3\)