Question

Find value of x, y
(√32)^x ÷2^y+1=1,16^4-x/2 -8^y=0
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Answer 1

Answer:

$\frac{ {(\sqrt{32})}^{x} }{ {2}^{y} + 1 } = 1 \\ {32}^{ \frac{x}{2} } = {2}^{y} + 1 \\ {2}^{ 5 \times \frac{x}{2} } = {2}^{y} + {2}^{0} \\ \blue{ \mathfrak{ \large \underline{on \: comparing} }}: \\ \frac{5x}{2} = y + 0 \\ 5x = 2y \\ x = \frac{2y}{5} - (i)\\ {16}^{ \frac{4 - x}{2} } - {8}^{y} = 0 \\ {16}^{ \frac{4 - x}{2} } = {8}^{y} \\ {2}^{4 \times \frac{(4 - x)}{2} } = {2}^{3y} \\ {2}^{(8 - 2x)} = {2}^{3y} \\ \blue{ \mathfrak{ \large\underline{on \: comparing }}}: \\ 8 - 2x = 3y - (ii) \\ \blue{ \mathfrak{ \large\underline{putting \: value \: of \: x \: in \: equation \: (ii)}}} : \\ 8 - 2 \times \frac{2y}{5} = 3y \\ 8 - \frac{4y}{5} = 3y \\ \frac{40 - 4y}{5} = 3y \\ 40 - 4y = 15y \\ 40 = 19y \\ y = \frac{40}{19} \\ \blue{ \mathfrak{ \large\underline{put \: the \: value \: of \: y \: in \: equation \: (i )}}}: \\ x = \frac{2 \times 40}{5 \times 19} \\ x = \frac{16}{19}$

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