Use the logarithm property a − b = (a)/(b) to combine the terms on the left side.

Recall that $\ln y = x$ is equivalent to $y = e^x$.

$$\frac{-3x-1}{7} = e^2$$

$$x = \frac{7e^2 + 1}{-3}$$ $$x = -\frac{7e^2 + 1}{3}$$

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.