1. **State the problem:** Solve the equation $\log_4(4x^2 + 7) - \log_4(x+1) = 2$ for $x$. 2. **Recall the logarithm subtraction rule:** $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$. 3. **Apply the rule:** $$\log_4 \left(\frac{4x^2 + 7}{x+1}\right) = 2$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\frac{4x^2 + 7}{x+1} = 4^2$$ 5. **Calculate $4^2$:** $$\frac{4x^2 + 7}{x+1} = 16$$ 6. **Multiply both sides by $(x+1)$ to clear the denominator:** $$\cancel{\frac{4x^2 + 7}{\cancel{x+1}}} \times \cancel{x+1} = 16(x+1)$$ $$4x^2 + 7 = 16x + 16$$ 7. **Bring all terms to one side to form a quadratic equation:** $$4x^2 + 7 - 16x - 16 = 0$$ $$4x^2 - 16x - 9 = 0$$ 8. **Simplify the quadratic equation:** $$4x^2 - 16x - 9 = 0$$ 9. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=4$, $b=-16$, and $c=-9$. 10. **Calculate the discriminant:** $$b^2 - 4ac = (-16)^2 - 4 \times 4 \times (-9) = 256 + 144 = 400$$ 11. **Calculate the roots:** $$x = \frac{-(-16) \pm \sqrt{400}}{2 \times 4} = \frac{16 \pm 20}{8}$$ 12. **Find the two possible solutions:** - $$x = \frac{16 + 20}{8} = \frac{36}{8} = 4.5$$ - $$x = \frac{16 - 20}{8} = \frac{-4}{8} = -0.5$$ 13. **Check for domain restrictions:** - The argument of the logarithm must be positive. - For $\log_4(4x^2 + 7)$, since $4x^2 + 7 > 0$ for all real $x$, no restriction here. - For $\log_4(x+1)$, we need $x + 1 > 0 \Rightarrow x > -1$. 14. **Check solutions against domain:** - $x=4.5$ satisfies $x > -1$. - $x=-0.5$ also satisfies $x > -1$. 15. **Final answer:** $$\boxed{x = 4.5 \text{ or } x = -0.5}$$