Question

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

x=3
x=1
x=0
x=6
x=11
x=9

Answer 1

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

**Step 1: Simplify the left side**

Using the properties of exponents, we know that:
\[
k^5 \cdot k^2 = k^{5 + 2} = k^7
\]
So the left side can be rewritten as:
\[
\frac{k^7}{k^x} = k^{7 - x}
\]

**Step 2: Simplify the right side**

The right side can be simplified using the property \((a^m)^n = a^{mn}\):
\[
(k^2)^{-2} = k^{2 \cdot (-2)} = k^{-4}
\]

**Step 3: Set the simplified left side equal to the simplified right side**
Now we can set the two sides equal:
\[
k^{7 - x} = k^{-4}
\]

Since the bases are the same, we can set the exponents equal to each other:
\[
7 - x = -4
\]

**Step 4: Solve for \(x\)**
To find \(x\):
\[
7 + 4 = x
\]
\[
x = 11
\]

So the missing exponent \(x\) that makes the equation true is:

\[
\boxed{11}
\]