Question

If 2^m+n÷2^m-n=16 and a=2^1÷10 then (a^2m+n-p)^2÷(a^m-2n+2p)^-1

Answer 1

Answer:

$\frac{(a^{2m+n-p})^2}{(a^{m -2n + 2p})^{-1}} = 2$

Step-by-step explanation:

If 2^m+n / 2^n-m = 16 & a = 2^1/10 then a^2m+n-p / (am-2n+2p)^-1

$\frac{2^{m+n}}{2^{n-m}} = 16\\ \\\implies 2^{m + n - n + m} = 2^4 \\ \\\implies 2^{2m} = 2^4 \\ \\\implies 2m = 4$

=> m = 2

$\frac{(a^{2m+n-p})^2}{(a^{m -2n + 2p})^{-1}} \\ \\\implies a^{2m+n-p} \times a^{2m+n-p} \times a^{m -2n + 2p} \\ \\\implies a^{2m + n - p + 2m + n - p + m - 2n + 2p} \\ \\\implies a^{5m} \\ \\m =2\\ \\\implies a^{10}$

$a = 2^{\frac{1}{10}}\\ \\\implies a^{10} = (2^{\frac{1}{10}})^{10} = 2$