To solve for \( x \) in the equation
\[ \frac{k^{-5} \cdot k^2}{k^x} = (k^0)^{-2}, \]
let's simplify both sides of the equation.
Starting with the left side:
\[ k^{-5} \cdot k^2 = k^{-5 + 2} = k^{-3}, \]
so we have
\[ \frac{k^{-3}}{k^x} = k^{-3 - x}. \]
Now, for the right side, since \( k^0 = 1 \), we find
\[ (k^0)^{-2} = 1^{-2} = 1 = k^0. \]
Now our equation simplifies to:
\[ k^{-3 - x} = k^0. \]
Since the bases are the same, we can set the exponents equal to each other:
\[ -3 - x = 0. \]
Now, solve for \( x \):
\[ -x = 3 \implies x = -3. \]
Thus, the missing exponent \( x \) that makes the equation true is
\[ \boxed{-3}. \]