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. Use the power law of logarithms to combine the terms.

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

3. Equate the arguments and form a quadratic equation.

$(2x-1)^2 = 4-5x$
$4x^2 - 4x + 1 = 4 - 5x$
$4x^2 + x - 3 = 0$

4. Solve the quadratic equation and reject any invalid solutions.

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

For logarithms to be defined, the base $x > 0$ and $x \neq 1$, and the arguments $2x-1 > 0$ and $4-5x > 0$.
When $x = -1$, the base is negative, so $x = -1$ is rejected.
When $x = \frac{3}{4}$, $2(\frac{3}{4})-1 = 0.5 > 0$ and $4-5(\frac{3}{4}) = 0.25 > 0$, so it is valid.

$x = \frac{3}{4}$