Question 2: \((vi)\) \(\quad\) \(\large{\frac{( 16)^{x+1} +20\left( 4^{2x}\right)}{2^{x-3} \times 8^{x+2}}}\)
Solution:
\[\begin{align*}
&=\frac{( 16)^{x+1} +20\left( 4^{2x}\right)}{2^{x-3} \times 8^{x+2}}\\
&=\frac{(2^4)^{x+1} +2^2\times 5\times 2^{2 \times 2x}}{2^{x-3} \times 2^{3(x+2)}}\\
&=\frac{2^{4x+4} +5\times 2^{4x+2}}{ 2^{x-3+3x+6}}\\
&=\frac{2^{4x+2} (2^2 +5)}{ 2^{4x+3}}\\
&=\frac{2^{4x+2} (4 +5)}{ 2^{4x+2}\times 2}\\
&=\frac{9}{2} \\
\end{align*}\]