Solve the equation $x + 1 + \sqrt{5 - x} = 0$.

1. Change the base of the logarithm on the right-hand side to $x$.

$\log_x (2x-1) = \frac{\log_x (4-5x)}{\log_x x^2}$
$\log_x (2x-1) = \frac{\log_x (4-5x)}{2}$

2. Rearrange and use logarithm laws to equate the arguments.

$2 \log_x (2x-1) = \log_x (4-5x)$
$\log_x (2x-1)^2 = \log_x (4-5x)$
$(2x-1)^2 = 4-5x$

3. Expand and solve the resulting quadratic equation.

$4x^2 - 4x + 1 = 4 - 5x$
$4x^2 + x - 3 = 0$
$(4x - 3)(x + 1) = 0$
$x = \frac{3}{4} \text{ or } x = -1$

4. Check the validity of the solutions against the conditions for logarithms.

For logarithms to be defined, the base $x > 0$ and $x \neq 1$, and the arguments must be positive.
If $x = -1$, the base is negative, so $x = -1$ is rejected.
If $x = \frac{3}{4}$, $2(\frac{3}{4})-1 = \frac{1}{2} > 0$ and $4-5(\frac{3}{4}) = \frac{1}{4} > 0$, which are valid.
Therefore, $x = \frac{3}{4}$.