Step 1: 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: $$\frac{1}{\log_5 x} + 4 \log_5 x = 4$$ Step 2: Introduce a substitution to simplify the equation. Let $y = \log_5 x$. The equation becomes: $$\frac{1}{y} + 4y = 4$$ Step 3: Solve the algebraic equation for $y$. 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 $y$: $$2y = 1$$ $$y = \frac{1}{2}$$ Step 4: Substitute back to find $x$. 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 $b^c = a$. $$x = 5^{1/2}$$ $$x = \sqrt{5}$$ Step 5: Check the solution. 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 $\boxed{\sqrt{5}}$. That's 2 down. 3 left today — send the next one.