What is the missing exponent that would make this equation true?
k^−5⋅k^2/k^x = (k^0)^−2 (1 point)
To solve the equation, we can simplify the expressions on both sides and equate them.
On the left side, we have k^(-5) * k^2 / k^x.
When multiplying with the same base, we add the exponents: k^(-5) * k^2 = k^(-5 + 2) = k^(-3).
Therefore, the expression on the left side becomes k^(-3) / k^x = k^(-3 - x).
On the right side, we have (k^0)^(-2).
Any number raised to the power of 0 is 1, so (k^0)^(-2) = 1^(-2) = 1.
Setting the expressions on both sides equal to each other, we have k^(-3 - x) = 1.
For any non-zero number raised to the power of 0, the result is always 1. So we can write this equation as k^(-3 - x) = k^0.
In order for two powers with the same base to be equal, their exponents must be equal.
Therefore, we have -3 - x = 0.
To find the missing exponent, we solve for x:
-3 - x = 0
-x = 3
x = -3
Therefore, the missing exponent that would make the equation true is -3.