1. **Problem 1(b): Solve the equation** $\log_2(x + 3) = 3 - \log_2(x + 2)$. 2. **Rewrite the equation:** Move all logarithmic terms to one side: $$\log_2(x + 3) + \log_2(x + 2) = 3$$ 3. **Use the logarithm product rule:** $$\log_2\big((x + 3)(x + 2)\big) = 3$$ 4. **Convert from logarithmic to exponential form:** $$ (x + 3)(x + 2) = 2^3 = 8 $$ 5. **Expand and simplify:** $$ x^2 + 2x + 3x + 6 = 8 $$ $$ x^2 + 5x + 6 = 8 $$ 6. **Bring all terms to one side:** $$ x^2 + 5x + 6 - 8 = 0 $$ $$ x^2 + 5x - 2 = 0 $$ 7. **Solve the quadratic equation using the quadratic formula:** $$ x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times (-2)}}{2 \times 1} = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2} $$ 8. **Check domain restrictions:** - Since $\log_2(x+3)$ and $\log_2(x+2)$ are defined only if $x+3 > 0$ and $x+2 > 0$, we require $x > -3$ and $x > -2$. - Both solutions must satisfy $x > -2$. 9. **Approximate solutions:** $$ x = \frac{-5 + \sqrt{33}}{2} \approx 0.372 \quad \text{(valid)} $$ $$ x = \frac{-5 - \sqrt{33}}{2} \approx -5.372 \quad \text{(invalid, outside domain)} $$ --- 10. **Problem 1(c): Find the time $t$ when the chemical is free of impurity, i.e., when** $$ A = 3e^{4t} - 7e^{2t} - 6 = 0 $$ 11. **Substitute $y = e^{2t}$ to simplify:** $$ 3y^2 - 7y - 6 = 0 $$ 12. **Solve the quadratic equation for $y$:** $$ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 3 \times (-6)}}{2 \times 3} = \frac{7 \pm \sqrt{49 + 72}}{6} = \frac{7 \pm \sqrt{121}}{6} = \frac{7 \pm 11}{6} $$ 13. **Calculate the two roots:** $$ y_1 = \frac{7 + 11}{6} = 3 $$ $$ y_2 = \frac{7 - 11}{6} = -\frac{2}{3} $$ 14. **Since $y = e^{2t} > 0$, discard $y_2 = -\frac{2}{3}$ (negative).** 15. **Solve for $t$ using $y_1 = 3$:** $$ e^{2t} = 3 $$ $$ 2t = \ln 3 $$ $$ t = \frac{\ln 3}{2} $$ 16. **Final answer:** $$ t = \frac{\ln 3}{2} \approx 0.5493 \text{ minutes} $$