Solve the following system of equations: $\begin{cases} \log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right) \\ x^2 + 7x - 8 = 0 \end{cases}$
Given system of equations:
$\log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right)$ $\;\;\; \cdots \; (1)$
$x^2 + 7x - 8 = 0$ $\;\;\; \cdots \; (2)$
Solving equation $(2)$ gives
$\left(x + 8\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = -8$ $\;$ or $\;$ $x = 1$
When $\;$ $x = -8$, $\;$ the term $\;$ $\log \left(x + 3\right)$ $\;$ in equation $(1)$ becomes $\;$ $\log \left(-8 + 3\right) = \log \left(-5\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $x = -8$ $\;$ is not a valid solution to the given pair of equations.
Substituting $\;$ $x = 1$ $\;$ in equation $(1)$ gives
$\log \sqrt{5 - 1} + \log 2 = \log \left(1 + 3\right)$
i.e. $\;$ $\log \sqrt{4} + \log 2 = \log 4$
i.e. $\;$ $\log 2 + \log 2 = \log 4$
i.e. $\;$ $2 \log 2 = \log 4$
i.e. $\;$ $\log 2^2 = \log 4$ $\implies$ $\log 4 = \log 4$
$\therefore \;$ $x = 1$ $\;$ satisfies both equations $(1)$ and $(2)$.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{1 \right\}$