Rewrite the equation using the change of base formula.

The given equation is:

$$\log_x 5 + 4 \log_5 x = 4$$

Use the change of base formula $\log_b a = \frac{1}{\log_a b}$.
So, $\log_x 5 = \frac{1}{\log_5 x}$.
Substitute this into the equation:

$$\frac{1}{\log_5 x} + 4 \log_5 x = 4$$

Let $y = \log_5 x$.
The equation becomes:

$$\frac{1}{y} + 4y = 4$$

Multiply the entire equation by $y$ to eliminate the fraction (note that $y \neq 0$ since $\log_5 x = 0$ implies $x=1$, which is not allowed as a base for $\log_x 5$):

$$1 + 4y^2 = 4y$$

Rearrange the terms to form a standard quadratic equation:

$$4y^2 - 4y + 1 = 0$$

This is a perfect square trinomial, which can be factored as $(2y - 1)^2$:

$$(2y - 1)^2 = 0$$

Take the square root of both sides:

$$2y - 1 = 0$$

Solve for $y$:

$$2y = 1$$ $$y = \frac{1}{2}$$

Recall that $y = \log_5 x$. Substitute the value of $y$ back:

$$\log_5 x = \frac{1}{2}$$

Convert the logarithmic equation to an exponential equation. If $\log_b a = c$, then $b^c = a$.

$$x = 5^{1/2}$$ $$x = \sqrt{5}$$

For $\log_x 5$ to be defined, $x > 0$ and $x \neq 1$.
For $\log_5 x$ to be defined, $x > 0$.
Our solution $x = \sqrt{5}$ satisfies these conditions, as $\sqrt{5} \approx 2.236$, which is positive and not equal to 1.

The final answer is $\sqrt{5}$.