Solve the equation: $\;$ $\log_3 \left(x + 1\right) + \log_3 \left(x + 3\right) = 1$
Given equation: $\;\;$ $\log_3 \left(x + 1\right) + \log_3 \left(x + 3\right) = 1$
i.e. $\;$ $\log_3 \left[\left(x + 1\right) \left(x + 3\right)\right] = 1$
$\implies$ By definition, $\;$ $x^2 + 4x + 3 = 3^1 = 3$
i.e. $\;$ $x^2 + 4x = 0$
i.e. $\;$ $x \left(x + 4\right) = 0$
i.e. $\;$ $x = 0$, $\;$ or $\;$ $x = -4$
When $\;$ $x = -4$, $\;$ the Left Hand Side of the given problem becomes
$\log_3 \left(-4 + 1\right) + \log_3 \left(-4 + 3\right) = \log_3 \left(-3\right) + \log_3 \left(-1\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $x = -4$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$