1. **State the problem:** Solve the equation $$\log_{x} 1000 - \log x^{2} + \left(10^{-3}\right)^{0} = 0$$ for $x$. 2. **Recall logarithm properties:** - Change of base formula: $\log_a b = \frac{\log b}{\log a}$ (logarithm base can be changed to any base, commonly 10 or $e$). - Power rule: $\log a^{b} = b \log a$. - $\left(10^{-3}\right)^{0} = 1$ since any nonzero number to the zero power is 1. 3. **Rewrite the terms:** - $\log_{x} 1000 = \frac{\log 1000}{\log x}$. - $\log x^{2} = 2 \log x$. 4. **Substitute and simplify:** $$\frac{\log 1000}{\log x} - 2 \log x + 1 = 0$$ 5. **Calculate $\log 1000$ (base 10):** $$\log 1000 = 3$$ 6. **Let $y = \log x$ to simplify:** $$\frac{3}{y} - 2y + 1 = 0$$ 7. **Multiply through by $y$ (assuming $y \neq 0$):** $$3 - 2y^{2} + y = 0$$ 8. **Rearrange into standard quadratic form:** $$-2y^{2} + y + 3 = 0$$ Multiply both sides by $-1$: $$2y^{2} - y - 3 = 0$$ 9. **Solve quadratic equation $2y^{2} - y - 3 = 0$ using quadratic formula:** $$y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^{2} - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}$$ 10. **Calculate roots:** - $y = \frac{1 + 5}{4} = \frac{6}{4} = 1.5$ - $y = \frac{1 - 5}{4} = \frac{-4}{4} = -1$ 11. **Recall $y = \log x$, so:** - For $y=1.5$: $\log x = 1.5 \implies x = 10^{1.5} = 10^{\frac{3}{2}} = \sqrt{10^{3}} = \sqrt{1000} \approx 31.622$ - For $y=-1$: $\log x = -1 \implies x = 10^{-1} = 0.1$ 12. **Check domain restrictions:** - Base $x$ of logarithm must be positive and not equal to 1. - $x \approx 31.622 > 0$ and $\neq 1$ valid. - $x = 0.1 > 0$ and $\neq 1$ valid. 13. **Final answer:** $$x = 10^{1.5} \approx 31.622 \quad \text{or} \quad x = 0.1$$