If $2^{n+m}/(2^{n-m}) = 16$ and $a = 2^{1/10}$, then, $(a^{2m+n-p})^2/(a^{m-2n+2p})^{-1} = ?$

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

$2^{(n+m)-(n-m)} = 16$

$2^{n+m-n+m} = 16$

$2^{2m} = 2^4$

$2m = 4$

$m = 2$

---

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

Dividing by something raised to $-1$ is the same as multiplying by it, so:

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

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

$= a^{4m+2n-2p} \cdot a^{m-2n+2p}$

$= a^{(4m+2n-2p)+(m-2n+2p)}$

$= a^{5m}$

All the $n$ and $p$ terms cancel out.

---

Substituting $m = 2$ and $a = 2^{1/10}$:

$a^{5m} = a^{5 \times 2} = a^{10}$

$= (2^{1/10})^{10}$

$= 2^{10/10}$

$= 2$