Here,
\(\frac{2^{x-1}+4\times2^x}{2^{x-1}}\)
= \(\frac{2^x\times{\displaystyle\frac12}+4\times2^x}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{2^x\left({\displaystyle\frac12}+4\right)}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{\displaystyle\frac{1+8}2}{\displaystyle\frac12}\)
= \(\frac92\times\frac21\)
= 9 is the required answer