1. **State the problem:** Solve the equation $$\ln\left(\frac{x^2+4}{4}\right) = \ln 4$$ for $x$. 2. **Recall the property of logarithms:** If $\ln A = \ln B$, then $A = B$, provided $A > 0$ and $B > 0$. 3. **Apply the property:** Set the arguments equal: $$\frac{x^2+4}{4} = 4$$ 4. **Multiply both sides by 4 to clear the denominator:** $$\cancel{4} \times \frac{x^2+4}{\cancel{4}} = 4 \times 4$$ $$x^2 + 4 = 16$$ 5. **Subtract 4 from both sides:** $$x^2 + 4 - 4 = 16 - 4$$ $$x^2 = 12$$ 6. **Take the square root of both sides:** $$x = \pm \sqrt{12}$$ 7. **Simplify the square root:** $$x = \pm \sqrt{4 \times 3} = \pm 2\sqrt{3}$$ 8. **Check the domain:** Since the original logarithm argument is $\frac{x^2+4}{4}$, and $x^2 + 4 > 0$ for all real $x$, both solutions are valid. **Final answer:** $$x = \pm 2\sqrt{3}$$