To solve the equation
\[ \frac{k^5 \cdot k^2}{k^x} = (k^2)^{-2}, \]
we will simplify both sides.
First, simplify the left-hand side:
\[ k^5 \cdot k^2 = k^{5 + 2} = k^7. \]
Now, the left side becomes:
\[ \frac{k^7}{k^x} = k^{7 - x}. \]
Next, simplify the right-hand side:
\[ (k^2)^{-2} = k^{2 \cdot (-2)} = k^{-4}. \]
Now we have the equation:
\[ k^{7 - x} = k^{-4}. \]
Since the bases (k) are the same, we can set the exponents equal to each other:
\[ 7 - x = -4. \]
To find \(x\), solve for it:
\[ 7 + 4 = x, \]
\[ x = 11. \]
Thus, the missing exponent is
\[ \boxed{11}. \]