Here is the solution to problem 24: 24) $\ln (-3x - 1) - \ln 7 = 2$ Step 1: Use the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$ to combine the terms on the left side. $$\ln \left(\frac{-3x-1}{7}\right) = 2$$ Step 2: Convert the logarithmic equation to an exponential equation. Recall that $\ln y = x$ is equivalent to $y = e^x$. $$\frac{-3x-1}{7} = e^2$$ Step 3: Multiply both sides by 7. $$-3x-1 = 7e^2$$ Step 4: Add 1 to both sides. $$-3x = 7e^2 + 1$$ Step 5: Divide by -3 to solve for $x$. $$x = \frac{7e^2 + 1}{-3}$$ $$x = -\frac{7e^2 + 1}{3}$$ Step 6: Check the domain. The argument of the logarithm, $-3x-1$, must be greater than 0. Substitute the value of $x$: $$-3\left(-\frac{7e^2 + 1}{3}\right) - 1$$ $$= (7e^2 + 1) - 1$$ $$= 7e^2$$ Since $e^2$ is positive, $7e^2$ is positive. Thus, the solution is valid. The exact answer is: $$\boxed{x = -\frac{7e^2 + 1}{3}}$$