What is the missing exponent that would make this equation true? k^5•k^2/k^x=(k^2)^-2

3 answers

To find the missing exponent that would make the equation true, we can simplify both sides of the equation using the exponent rules.

First, let's simplify the left side of the equation:
k^5 * k^2 / k^x can be simplified as k^(5 + 2 - x).

Now, let's simplify the right side of the equation:
(k^2)^-2 can be rewritten as k^(2*(-2)) = k^(-4).

Therefore, we have the equation k^(5 + 2 - x) = k^(-4).

Since the bases are the same on both sides of the equation (k), we can set the exponents equal to each other:
5 + 2 - x = -4.

Combining like terms, we get:
7 - x = -4.

To isolate x, we subtract 7 from both sides:
-x = -11.

Finally, we multiply both sides by -1 to solve for x:
x = 11.

Therefore, the missing exponent that would make the equation true is x = 11.
Are you sure explain
Apologies for the incorrect explanation. Here's the correct explanation:

To find the missing exponent that would make the equation true, let's simplify both sides of the equation using the exponent rules.

On the left side:
k^5 * k^2 / k^x can be simplified as k^(5 + 2 - x).

On the right side:
(k^2)^-2 is equal to 1/(k^2)^2, which simplifies to 1/k^4.

Therefore, we have the equation k^(5 + 2 - x) = 1/k^4.

Since the bases are the same on both sides of the equation (k), we can set the exponents equal to each other:

5 + 2 - x = -4.

Combining like terms, we get:

7 - x = -4.

To isolate x, we subtract 7 from both sides:

-x = -11.

Multiplying both sides by -1:

x = 11.

Therefore, the missing exponent that would make the equation true is x = 11.