Question

What is the missing exponent that would make this equation true? k to the power of negative five times k to the power of two divided by k to the power of x equals (k to the power of zero) to the power of negative two

Answer 1

To solve the equation \( \frac{k^{-5} \cdot k^{2}}{k^{x}} = (k^{0})^{-2} \), we need to simplify both sides of the equation.

First, let's simplify the left-hand side:

1. Combine the exponents in the numerator:
\[
k^{-5} \cdot k^{2} = k^{-5 + 2} = k^{-3}
\]

2. Now substitute back into the equation:
\[
\frac{k^{-3}}{k^{x}} = k^{-3 - x}
\]

Now, we have:
\[
k^{-3 - x}
\]

Next, let's simplify the right-hand side:

1. The expression \( (k^{0})^{-2} \):
- Since \( k^{0} = 1 \), we have:
\[
(k^{0})^{-2} = 1^{-2} = 1
\]

Now, we can equate both sides:
\[
k^{-3 - x} = 1
\]

Since \( k^{0} = 1 \), we can also say that:
\[
-3 - x = 0
\]

Now, we solve for \( x \):
\[
-3 - x = 0
\]
\[
-x = 3
\]
\[
x = -3
\]

Thus, the missing exponent \( x \) that would make the equation true is:
\[
\boxed{-3}
\]