Solve for k in the equation \(\frac{1}{8}^{k+2}\) = 1
- 2
- -4
-
-2
- 4
The correct answer is: C
Explanation
\((8^{-1})^{k+2}\) = \((8^{0})\)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
There is an explanation video available .