Solve for k in the equation \(\frac{1}{8}^{k+2}\) = 1
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Correct Answer: Option C
Explanation:
\((8^{-1})^{k+2}\) =\((8^{0})\)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
\((8^{-1})^{k+2}\) =\((8^{0})\)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2