What is the missing exponent that would make this equation true?

Question

k^−5 ⋅ k^2/k^x = (k^0)^−2

Answer 1

To solve the equation

\[ \frac{k^{-5} \cdot k^2}{k^x} = (k^0)^{-2}, \]

let's simplify both sides step by step.

1. Simplify the right-hand side: Since \(k^0 = 1\), we have: \[ (k^0)^{-2} = 1^{-2} = 1. \]

2. Now simplify the left-hand side: Using the properties of exponents, calculate \(k^{-5} \cdot k^2\): \[ k^{-5} \cdot k^2 = k^{-5 + 2} = k^{-3}. \] Therefore, the left-hand side becomes: \[ \frac{k^{-3}}{k^x} = k^{-3 - x}. \]

Now, we can rewrite the equation as: \[ k^{-3 - x} = 1. \]

3. Solve for the exponent: Since \(1\) can be rewritten as \(k^0\) (as long as \(k \neq 0\)), we equate the exponents: \[ -3 - x = 0. \]

4. Isolate \(x\): \[ -x = 3 \quad \Longrightarrow \quad x = -3. \]

Thus, the missing exponent \(x\) that would make the equation true is

\[ \boxed{-3}. \]