To solve the equation \( \frac{k^5 \cdot k^2}{k^x} = (k^2)^{-2} \), we can start by simplifying both sides.
Step 1: Simplify the left side
Using the properties of exponents, we know that: \[ k^5 \cdot k^2 = k^{5 + 2} = k^7 \] So the left side can be rewritten as: \[ \frac{k^7}{k^x} = k^{7 - x} \]
Step 2: Simplify the right side
The right side can be simplified using the property \((a^m)^n = a^{mn}\): \[ (k^2)^{-2} = k^{2 \cdot (-2)} = k^{-4} \]
Step 3: Set the simplified left side equal to the simplified right side Now we can set the two sides equal: \[ k^{7 - x} = k^{-4} \]
Since the bases are the same, we can set the exponents equal to each other: \[ 7 - x = -4 \]
Step 4: Solve for \(x\) To find \(x\): \[ 7 + 4 = x \] \[ x = 11 \]
So the missing exponent \(x\) that makes the equation true is:
\[ \boxed{11} \]