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

k^5 ⋅ k^2/k^x = (k^2)^−2

1 answer

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

\[ \boxed{11}. \]