If \(2^n = 128\), find the value of \(2^{n – 1})(5^{n – 2})\)

Explanation

\(2^n\) = 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(\(10^5\))