The equation given is:

$\log_x 1000 - \log_x x^2 + (10^{-3})^0 = 0$

Let's solve it step-by-step:

1. Simplify $(10^{-3})^0$: $(10^{-3})^0 = 1$

2. Apply the property of logarithms $\log_a b^c = c\log_a b$: $\log_x 1000 - 2\log_x x + 1 = 0$ Recall that $\log_x x = 1$, so: $\log_x 1000 - 2 \cdot 1 + 1 = 0$ Simplifying further: $\log_x 1000 - 2 + 1 = 0 \quad \Rightarrow \quad \log_x 1000 - 1 = 0$

3. Solving for $\log_x 1000 = 1$: The equation $\log_x 1000 = 1$ means that $x^1 = 1000$. Therefore, $x = 1000$.

So, the solution to the equation is:

$x = 1000$

Would you like me to go into more detail, or do you have any further questions?

Relative Questions:

1. What are the properties of logarithms that are most useful in simplifying equations?
2. How do you interpret logarithmic equations with different bases?
3. Can this equation have any other solution if interpreted differently?
4. How do you change the base of a logarithm for solving equations?
5. What are some other techniques for solving logarithmic equations?

Tip: Always check if you can simplify logarithmic expressions using properties like the change of base or power rules to reduce the complexity of the problem.