WAEC - Mathematics (1999 - No. 16)
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
\(4.5\times 2^{n}\)
\(4.5\times 2^{2n}\)
\(9\times 2^{n-1}\)
\(9\times 2^{2n}\)
Explication
\(K = 81 \times 2^{2n - 2}\)
\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)
= \(9 \times 2^{n - 1}\)