Let's solve the given equation:

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

### Step 1: Simplify the expression \((10^{-3})^0\)
\((10^{-3})^0\) is equal to 1 because any non-zero number raised to the power of 0 is 1.

So, the equation becomes:

\[
\log_x 1000 - \log_{10} (x^2) + 1 = 0
\]

### Step 2: Rearrange the Equation
Subtract 1 from both sides:

\[
\log_x 1000 - \log_{10} (x^2) = -1
\]

### Step 3: Convert Logarithms to a Common Base
We have two different logarithms here: \(\log_x 1000\) and \(\log_{10} (x^2)\). Let's express \(\log_x 1000\) in terms of base 10.

Using the change of base formula for logarithms:

\[
\log_x 1000 = \frac{\log_{10} 1000}{\log_{10} x}
\]

And we already know that:

\[
\log_{10} 1000 = 3
\]

Thus:

\[
\log_x 1000 = \frac{3}{\log_{10} x}
\]

Similarly, \(\log_{10} (x^2) = 2 \cdot \log_{10} x\).

### Step 4: Substitute and Solve
Now substitute these back into the equation:

\[
\frac{3}{\log_{10} x} - 2 \cdot \log_{10} x = -1
\]

### Step 5: Set \(y = \log_{10} x\)
To simplify the equation, let \(y = \log_{10} x\). The equation becomes:

\[
\frac{3}{y} - 2y = -1
\]

### Step 6: Clear the Fraction
Multiply through by \(y\) (assuming \(y \neq 0\)):

\[
3 - 2y^2 = -y
\]

Rearrange to form a standard quadratic equation:

\[
2y^2 - y - 3 = 0
\]

### Step 7: Solve the Quadratic Equation
To solve \(2y^2 - y - 3 = 0\), use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 2\), \(b = -1\), and \(c = -3\). Substitute these values:

\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}
\]

\[
y = \frac{1 \pm \sqrt{1 + 24}}{4}
\]

\[
y = \frac{1 \pm \sqrt{25}}{4}
\]

\[
y = \frac{1 \pm 5}{4}
\]

This gives us two possible solutions:

1. \(y = \frac{6}{4} = 1.5\)
2. \(y = \frac{-4}{4} = -1\)

### Step 8: Find \(x\)
Recall that \(y = \log_{10} x\).

#### Case 1: \(y = 1.5\)
\[
\log_{10} x = 1.5 \implies x = 10^{1.5} = \sqrt{1000} = 31.62
\]

#### Case 2: \(y = -1\)
\[
\log_{10} x = -1 \implies x = 10^{-1} = 0.1
\]

### Step 9: Final Answer
The two possible values of \(x\) are:

\[
x = 31.62 \quad \text{or} \quad x = 0.1
\]

Would you like further details or have any questions?

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Here are 5 related questions to expand on this:

1. How does the change of base formula for logarithms work?
2. What are some real-world applications of logarithmic equations?
3. How can we solve equations involving different bases of logarithms?
4. What does it mean for a number to have a logarithm in a particular base?
5. How do we interpret negative solutions in logarithmic equations?

**Tip**: Always check whether each possible value of \(x\) satisfies the domain requirements for logarithms, as logarithms are only defined for positive values.

This detailed solution walks through solving the logarithmic equation logx 1000 - log10(x^2) = 0 using the change of base formula and quadratic equation techniques. Suitable for students in Grades 9-12, this solution explains key concepts in logarithms and quadratic equations.

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