1. **State the problem:** Solve the equation $\log_4(4xe^{2} + 7) - \log_4(x+1) = 2$ for $x$. 2. **Recall the logarithm subtraction rule:** $\log_a A - \log_a B = \log_a \left(\frac{A}{B}\right)$. 3. **Apply the rule:** $$\log_4 \left(\frac{4xe^{2} + 7}{x+1}\right) = 2$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\frac{4xe^{2} + 7}{x+1} = 4^{2}$$ 5. **Calculate $4^{2}$:** $$4^{2} = 16$$ 6. **Set up the equation:** $$\frac{4xe^{2} + 7}{x+1} = 16$$ 7. **Multiply both sides by $(x+1)$ to clear the denominator:** $$4xe^{2} + 7 = 16(x+1)$$ 8. **Expand the right side:** $$4xe^{2} + 7 = 16x + 16$$ 9. **Bring all terms to one side:** $$4xe^{2} - 16x + 7 - 16 = 0$$ 10. **Simplify constants:** $$4xe^{2} - 16x - 9 = 0$$ 11. **Factor out $x$ from the terms with $x$:** $$x(4e^{2} - 16) - 9 = 0$$ 12. **Isolate $x$:** $$x(4e^{2} - 16) = 9$$ 13. **Divide both sides by $(4e^{2} - 16)$:** $$x = \frac{9}{\cancel{4e^{2} - 16}}$$ 14. **Check domain restrictions:** - $x+1 > 0 \Rightarrow x > -1$ - $4xe^{2} + 7 > 0 \Rightarrow x > -\frac{7}{4e^{2}}$ Since $-\frac{7}{4e^{2}} > -1$, the domain is $x > -\frac{7}{4e^{2}}$. 15. **Evaluate denominator:** $$4e^{2} - 16 = 4(e^{2} - 4)$$ Since $e^{2} \approx 7.389$, $e^{2} - 4 \approx 3.389 > 0$, denominator is positive. 16. **Final solution:** $$x = \frac{9}{4e^{2} - 16}$$ This $x$ satisfies the domain restrictions. **Answer:** $$\boxed{x = \frac{9}{4e^{2} - 16}}$$