Question

Please answer the attached question ​

Answer 1

`\frac{ {4}^(1 - y) {16}^(y - 1) }{ {8}^(y + 2) } = \frac{ {2}^(2(1 - y)) {2}^(4(y - 1)) }{ {2}^(3(y + 2)) } = \frac{ {2}^(2(1 - y) + 4(y - 1)) }{ {2}^(3(y + 2)) } = \frac{ {2}^(2 - 2y + 4y - 4) }{ {2}^(3y + 6) } = \frac{ {2}^(2y - 2) }{ {2}^(3y + 6) } = {2}^( - y - 8) = \frac{1}{ {2}^(y + 8) }`