1. **State the problem:** Solve the equation $$4 \log(x) = 2 \log(x) + \log(4) + 2$$ for $x$. 2. **Recall logarithm properties:** - $a \log(b) = \log(b^a)$ - $\log(a) + \log(b) = \log(ab)$ 3. **Rewrite the equation:** $$4 \log(x) - 2 \log(x) = \log(4) + 2$$ 4. **Simplify the left side:** $$\cancel{4 \log(x)} - \cancel{2 \log(x)} = 2 \log(x)$$ So the equation becomes: $$2 \log(x) = \log(4) + 2$$ 5. **Express $2$ as a logarithm:** Since $2 = \log(10^2) = \log(100)$ (assuming base 10), rewrite: $$2 \log(x) = \log(4) + \log(100)$$ 6. **Combine right side logarithms:** $$\log(4) + \log(100) = \log(4 \times 100) = \log(400)$$ 7. **Now the equation is:** $$2 \log(x) = \log(400)$$ 8. **Rewrite left side using power rule:** $$\log(x^2) = \log(400)$$ 9. **Since logarithms are equal, their arguments are equal:** $$x^2 = 400$$ 10. **Solve for $x$:** $$x = \pm \sqrt{400} = \pm 20$$ 11. **Check domain:** Logarithm is defined only for $x > 0$, so discard $x = -20$. **Final answer:** $$\boxed{20}$$