Question 1 Report
Solve for k in the equation \( \left(\frac{1}{8}\right)^{k+2} = 1 \)
2
-4
-2
4
Answer Details
(8?1)k+2 ( 8 ? 1 ) ? + 2 = (80) ( 8 0 )
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
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