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

(k^5 time k^2)/(k^x) = (k^2)^-2

A: x = 1
B: x = 6
C: x = 9
D: x = 11
E: x = 3
F: x = 0

1 answer

We can simplify the left side of the equation first:

(k^5)(k^2) / (k^x) = k^(5+2) / k^x = k^7 / k^x

Using the rule of exponents that states a^m / a^n = a^(m-n), we have:

k^7 / k^x = k^(7-x)

Now we can rewrite the equation as:

k^(7-x) = (k^2)^-2 = k^(-4)

For the equation to be true, the exponents on both sides of the equation must be equal. This means:

7 - x = -4

Solving for x, we find:

x = 7 + 4 = 11

Therefore, the missing exponent that would make the equation true is x = 11. The correct answer is D: x = 11.