Question

16×2^n+1-8×^2n/ 16×2^n+1-4×2^n+1 find the value of n solve it please.​

Answer 1

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$\: \: \: \: \: \tt{ \frac{16 \times 2 {}^{n + 1} - 8 \times2 {}^{n} }{16 \times 2 {}^{n + 1} - 4 \times 2 {}^{n + 1} } } \\ \\ \tt{= \frac{2 {}^{4} \times2 {}^{n + 1} - 2 {}^{3} \times 2 {}^{n} }{2 {}^{4} \times 2 {}^{n + 1} - 2 {}^{2} \times 2 {}^{n + 1} } } \\ \\ \tt{ = \frac{2 {}^{4 + (n + 1) }- 2 {}^{3 + n} }{2 {}^{4 + (n + 1)} - 2 {}^{2 + (n + 1)} }} \\ \\ \tt{ = \frac{2 {}^{4 + n + 1} - 2 {}^{3 + n} }{2 {}^{4 + n + 1} - 2 {}^{2 + n + 1}} } \\ \\ \tt{ = \frac{ \cancel{2 {}^{5 + n} - 2 {}^{3 + n}} }{ \cancel{2 {}^{5 + n} - 2 {}^{3 + n} } } } \\ \\ \tt{ = 1}$

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathbb{ \: \: SOLUTION \: \: } \mid}}}}$

$\: \: \: \bf{ \frac{16 \times 2 {}^{n + 1} - 8 \times2 {}^{n} }{16 \times 2 {}^{n + 1} - 4 \times 2 {}^{n + 1} } } \\ \\ \bf{= \frac{2 {}^{4} \times2 {}^{n + 1} - 2 {}^{3} \times 2 {}^{n} }{2 {}^{4} \times 2 {}^{n + 1} - 2 {}^{2} \times 2 {}^{n + 1} } } \\ \\ \bf{ = \frac{2 {}^{4 + (n + 1) }- 2 {}^{3 + n} }{2 {}^{4 + (n + 1)} - 2 {}^{2 + (n + 1)} }} \\ \\ \bf{ = \frac{2 {}^{4 + n + 1} - 2 {}^{3 + n} }{2 {}^{4 + n + 1} - 2 {}^{2 + n + 1}} } \\ \\ \bf{ = \frac{ \cancel{2 {}^{5 + n} - 2 {}^{3 + n}} }{ \cancel{2 {}^{5 + n} - 2 {}^{3 + n} } } } \\ \\ \bf{ = 1}$