1. **State the problem:** Solve the equation $\log_2 x = \log_4 (x+6)$.\n\n2. **Recall the change of base formula and properties of logarithms:**\nWe know that $\log_a b = \frac{\log_c b}{\log_c a}$ for any positive $c \neq 1$. Also, if $\log_a M = \log_a N$, then $M = N$ provided the bases and arguments are valid.\n\n3. **Rewrite $\log_4 (x+6)$ in terms of base 2:**\nSince $4 = 2^2$, we have\n$$\log_4 (x+6) = \frac{\log_2 (x+6)}{\log_2 4} = \frac{\log_2 (x+6)}{2}.$$\n\n4. **Substitute back into the equation:**\n$$\log_2 x = \frac{\log_2 (x+6)}{2}.$$\nMultiply both sides by 2 to clear the denominator:\n$$2 \log_2 x = \log_2 (x+6).$$\n\n5. **Use logarithm power rule:**\n$$\log_2 (x^2) = \log_2 (x+6).$$\n\n6. **Since the logs are equal and base 2 is the same, set the arguments equal:**\n$$x^2 = x + 6.$$\n\n7. **Rearrange to form a quadratic equation:**\n$$x^2 - x - 6 = 0.$$\n\n8. **Factor the quadratic:**\n$$ (x - 3)(x + 2) = 0.$$\n\n9. **Solve for $x$:**\n$$x = 3 \quad \text{or} \quad x = -2.$$\n\n10. **Check domain restrictions:**\nSince $\log_2 x$ is defined only for $x > 0$, $x = -2$ is invalid.\n\n11. **Final solution:**\n$$\boxed{3}.$$