1. Problem 49: Solve for $y$ given $\ln y = 2t + 4$. 2. Recall that if $\ln y = A$, then $y = e^A$. 3. Applying this, $y = e^{2t + 4} = e^4 \cdot e^{2t}$. 4. Problem 50: Solve for $y$ given $\ln y = -t + 5$. 5. Using the same rule, $y = e^{-t + 5} = e^5 \cdot e^{-t}$. 6. Problem 51: Solve for $y$ given $\ln(y - b) = 5t$. 7. Exponentiate both sides: $y - b = e^{5t}$. 8. Therefore, $y = b + e^{5t}$. 9. Problem 52: Solve for $y$ given $\ln(c - 2y) = t$. 10. Exponentiate: $c - 2y = e^t$. 11. Rearranging: $2y = c - e^t$. 12. So, $y = \frac{c - e^t}{2}$. 13. Problem 53: Solve for $y$ given $\ln(y - 1) - \ln 2 = x + \ln x$. 14. Use log subtraction rule: $\ln \frac{y - 1}{2} = x + \ln x$. 15. Rewrite right side: $x + \ln x = \ln(e^x) + \ln x = \ln(x e^x)$. 16. So, $\ln \frac{y - 1}{2} = \ln(x e^x)$. 17. Exponentiate both sides: $\frac{y - 1}{2} = x e^x$. 18. Multiply both sides by 2: $y - 1 = 2 x e^x$. 19. Finally, $y = 1 + 2 x e^x$. 20. Problem 54: Solve for $y$ given $\ln(y^2 - 1) - \ln(y + 1) = \ln(\sin x)$. 21. Use log subtraction rule: $\ln \frac{y^2 - 1}{y + 1} = \ln(\sin x)$. 22. Note $y^2 - 1 = (y - 1)(y + 1)$, so $\frac{y^2 - 1}{y + 1} = y - 1$. 23. So, $\ln(y - 1) = \ln(\sin x)$. 24. Exponentiate: $y - 1 = \sin x$. 25. Therefore, $y = 1 + \sin x$. Final answers: 49. $y = e^4 e^{2t}$ 50. $y = e^5 e^{-t}$ 51. $y = b + e^{5t}$ 52. $y = \frac{c - e^t}{2}$ 53. $y = 1 + 2 x e^x$ 54. $y = 1 + \sin x$