Question

what number would make

k^-5 x k^2 / k^x true to (k^0)^-2?

Answer 1

To determine the value of \( x \) that makes the equation

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

true, we start by simplifying both sides of the equation.

### Step 1: Simplify the Left Side

Using the properties of exponents, we can combine the terms in the numerator:

\[
k^{-5} \cdot k^{2} = k^{-5 + 2} = k^{-3}
\]

Then, we can write the left side as:

\[
\frac{k^{-3}}{k^{x}} = k^{-3 - x}
\]

### Step 2: Simplify the Right Side

Now, we simplify the right side:

\[
(k^0)^{-2}
\]

Since \( k^0 = 1 \) for any non-zero \( k \), we have:

\[
(1)^{-2} = 1
\]

Thus, the right side simplifies to:

\[
1 = k^0
\]

### Step 3: Set the Exponents Equal

Now we set the simplified left side equal to the simplified right side:

\[
k^{-3 - x} = k^0
\]

Since the bases are the same, we can equate the exponents:

\[
-3 - x = 0
\]

### Step 4: Solve for \( x \)

Now, solve for \( x \):

\[
-3 - x = 0 \implies -x = 3 \implies x = -3
\]

### Conclusion

The value of \( x \) that satisfies the equation is

\[
\boxed{-3}
\]