Question

What is the missing exponent that would make this equation true?k5⋅k2/k^x = (k^2)^−2 (1 point)Responsesx = 11x = 11x = 6x = 6x = 3x = 3x = 9x = 9x = 1x = 1x = 0

Answer 1

To solve the equation \( \frac{k^5 \cdot k^2}{k^x} = (k^2)^{-2} \), we start simplifying both sides.

### Step 1: Simplify the left-hand side
\[
\frac{k^5 \cdot k^2}{k^x} = \frac{k^{5+2}}{k^x} = \frac{k^7}{k^x} = k^{7-x}
\]

### Step 2: Simplify the right-hand side
The right-hand side can be simplified as follows:
\[
(k^2)^{-2} = k^{2 \cdot -2} = k^{-4}
\]

### Step 3: Set the exponents equal
Now we can set the left-hand side equal to the right-hand side:
\[
k^{7-x} = k^{-4}
\]

Since the bases are the same (assuming \( k \neq 0 \)), we can equate the exponents:
\[
7 - x = -4
\]

### Step 4: Solve for \( x \)
Now we solve for \( x \):
\[
7 + 4 = x \implies x = 11
\]

Thus, the missing exponent is **\( x = 11 \)**.