1. **State the problem:** Solve the equation $\lg(x^2) - \lg^2(-x) = 3$ for $x$. 2. **Recall the definitions and domain restrictions:** - $\lg$ denotes the base-10 logarithm. - The argument of a logarithm must be positive. - So, for $\lg(x^2)$, since $x^2 \geq 0$, the argument is positive for all $x \neq 0$. - For $\lg(-x)$, the argument $-x$ must be positive, so $-x > 0 \Rightarrow x < 0$. 3. **Rewrite the equation:** $$\lg(x^2) - \left(\lg(-x)\right)^2 = 3$$ 4. **Use logarithm properties:** Since $\lg(x^2) = 2\lg|x|$, and for $x<0$, $|x| = -x$, so: $$2\lg(-x) - \left(\lg(-x)\right)^2 = 3$$ 5. **Let $y = \lg(-x)$, then the equation becomes:** $$2y - y^2 = 3$$ 6. **Rewrite as a quadratic equation:** $$-y^2 + 2y - 3 = 0$$ Multiply both sides by $-1$: $$y^2 - 2y + 3 = 0$$ 7. **Solve the quadratic equation:** The discriminant is: $$\Delta = (-2)^2 - 4 \times 1 \times 3 = 4 - 12 = -8$$ Since $\Delta < 0$, there are no real solutions for $y$. 8. **Conclusion:** No real $y$ satisfies the equation, so no real $x$ satisfies the original equation under the domain restrictions. **Final answer:** No real solution for $x$.