1. **Problem:** Solve for $x$ in the equation $\log_2(x^2 - 4) = 3$.\n\n2. **Formula and rules:** Recall that $\log_b(a) = c$ means $b^c = a$. Also, the argument of a logarithm must be positive, so $x^2 - 4 > 0$.\n\n3. **Step 1: Write the equation in exponential form:**\n$$\log_2(x^2 - 4) = 3 \implies x^2 - 4 = 2^3$$\n\n4. **Step 2: Simplify the right side:**\n$$x^2 - 4 = 8$$\n\n5. **Step 3: Solve for $x^2$:**\n$$x^2 = 8 + 4 = 12$$\n\n6. **Step 4: Find $x$:**\n$$x = \pm \sqrt{12} = \pm 2\sqrt{3}$$\n\n7. **Step 5: Check domain restrictions:**\nSince $x^2 - 4 > 0$,\n$$x^2 > 4 \implies |x| > 2$$\nBoth $2\sqrt{3} \approx 3.464$ and $-2\sqrt{3} \approx -3.464$ satisfy this.\n\n**Final answer:**\n$$x = \pm 2\sqrt{3}$$