Given: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}}$
To find: $\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}$
Find: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16$
By using rational components $\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}$ We get
$2^{\text{m}+\text{n}-\text{n}+\text{m}}=16$
$2^{2\text{m}}=2^4$
By equating rational exponents we get
$2\text{m}=4$
$\text{m}=\frac{4}{2}$
$\text{m}=2$
Now, $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})}=(\text{a}^{2\text{m}+\text{n}-\text{p}}).(\text{a}^{\text{m}-2\text{n}+2\text{p})}$ we get
$=\text{a}^{2\text{mn}+\text{n}+\text{p}+\text{m}-2\text{n}+2\text{p}}$
$=\text{a}^{3\text{m}-\text{n}+\text{p}}$
Now putting value of $\text{a}=2^\frac{1}{10}$ we get,
$=2^{\frac{3\text{m}-\text{n}+\text{p}}{10}}$
$=2^{\frac{6-\text{n}+\text{p}}{10}}$
Also, $\frac{3^\text{p}}{3^\text{n}}=81$
$3^{\text{p}-\text{n}}=3^4$
On comparing $LHS$ and $RHS$ we get, $p - n = 4.$
Now,
$\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=\text{a}^{3\text{m}-\text{n}+\text{p}}$
$=2^{\frac{6+(\text{p}-\text{n})}{10}}$
$=2^{\frac{6+4}{10}}$
$=2^{\frac{10}{10}}=2^1$
$=2$
So, option $(a)$ is the correct answer.
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