(vi) \( \frac{(16)^{x+1}+20\left(4^{2 x}\right)}{2^{x-3} \times 8^{x+2}} \)

To simplify the expression \( \frac{(16)^{x+1}+20\left(4^{2 x}\right)}{2^{x-3} \times 8^{x+2}} \), let's start by rewriting the base numbers in terms of powers of 2. 1. We know that \( 16 = 2^4 \) and \( 4 = 2^2 \). - Therefore, \( (16)^{x+1} = (2^4)^{x+1} = 2^{4(x+1)} = 2^{4x + 4} \). - Also, \( 4^{2x} = (2^2)^{2x} = 2^{4x} \). - Thus, \( 20(4^{2x}) = 20 \cdot 2^{4x} = 20 \cdot 2^{4x} = 20 \cdot 2^{4x} = 20 \cdot 2^{4x} \). 2. Substitute these results into the numerator: \[ (16)^{x+1} + 20(4^{2x}) = 2^{4x + 4} + 20 \cdot 2^{4x} \] We can factor \( 2^{4x} \) from both terms: \[ = 2^{4x}(2^4 + 20) = 2^{4x}(16 + 20) = 2^{4x}(36) \] 3. Now for the denominator, \( 2^{x-3} \times 8^{x+2} \). - Since \( 8 = 2^3 \), \( 8^{x+2} = (2^3)^{x+2} = 2^{3(x+2)} = 2^{3x + 6} \). - Thus, the denominator becomes: \[ 2^{x-3} \times 2^{3x+6} = 2^{(x-3) + (3x+6)} = 2^{4x + 3} \] 4. Now substitute both parts back into the original expression: \[ \frac{2^{4x} \cdot 36}{2^{4x + 3}} \] 5. Simplifying further: \[ = \frac{36}{2^{4x + 3 - 4x}} = \frac{36}{2^3} = \frac{36}{8} = 4.5 \] Thus, the simplified expression is \( 4.5 \).