If \(2^n = 128\), find the value of \(2^{n - 1})(5^{n - 2})\)
\(2^n\) = 128 \(2^n = 2^7\) n = 7 (2\(^{n - 1}\))(5\(^{n - 2}\)), then, put n = 7
= (2\(^{7 - 1}\))(5\(^{7 - 2}\))
= (\(2^6\))(\(2^5\))
= \(2^1 \times 2^5 \times 2^5\)
= 2 x (\(2 \times 5)^5\)
= 2(\(10^5\))
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