Certainly! Let's simplify the given expression step-by-step:
We are starting with the expression:
[tex]\[
\left( \frac{\left(x^2 y^3\right)^{-1}}{\left(x^{-2} y^2 z\right)^2} \right)^2
\][/tex]
First, simplify the inner parts separately.
### Step 1: Simplify [tex]\(\left(x^2 y^3\right)^{-1}\)[/tex]
[tex]\[
(x^2 y^3)^{-1} = \frac{1}{x^2 y^3}
\][/tex]
### Step 2: Simplify [tex]\(\left(x^{-2} y^2 z\right)^2\)[/tex]
[tex]\[
(x^{-2} y^2 z)^2 = (x^{-2})^2 (y^2)^2 (z)^2 = x^{-4} y^4 z^2
\][/tex]
### Step 3: Combining the simplified results
Next, we will insert these simplified results into the fraction:
[tex]\[
\frac{\left(x^2 y^3\right)^{-1}}{\left(x^{-2} y^2 z\right)^2} = \frac{\frac{1}{x^2 y^3}}{x^{-4} y^4 z^2} = \frac{1}{x^2 y^3} \cdot \frac{1}{x^{-4} y^4 z^2}
\][/tex]
[tex]\[
= \frac{1}{x^2 y^3} \cdot \frac{1}{x^{-4} y^4 z^2} = \frac{1}{x^2 y^3 x^{-4} y^4 z^2}
= \frac{1}{x^{2-4} y^{3+4} z^2} = \frac{1}{x^{-2} y^7 z^2}
\][/tex]
Since we want to express this with positive exponents:
[tex]\[
\frac{1}{x^{-2} y^7 z^2} = x^2 y^{-7} z^{-2}
\][/tex]
### Step 4: Raise to the power of 2
[tex]\[
(x^2 y^{-7} z^{-2})^2 = (x^2)^2 (y^{-7})^2 (z^{-2})^2 = x^{2 \cdot 2} y^{-7 \cdot 2} z^{-2 \cdot 2} = x^4 y^{-14} z^{-4}
\][/tex]
To express this with positive exponents:
[tex]\[
x^4 y^{-14} z^{-4} = x^4 \cdot \frac{1}{y^{14}} \cdot \frac{1}{z^4} = x^4 y^{-14} z^{-4}
\][/tex]
So, the simplified form of the original expression with positive exponents is:
[tex]\[
x^4 y^{-14} z^{-4}
\][/tex]
### Conclusion
- The exponent on [tex]\(x\)[/tex] is [tex]\(4\)[/tex].
- The exponent on [tex]\(y\)[/tex] is [tex]\(-14\)[/tex].
- The exponent on [tex]\(z\)[/tex] is [tex]\(-4\)[/tex].
