Question

If $\frac{2^{m+n}}{2^{n-m}}=16$, $\frac{3^p}{3^n}=81$ and $a=2^{1/10}$, then$\frac{a^{2m+n-p}}{(a^{m-2n+2p}{)}^{-1}}=$

(a) 2

(b) $\frac{1}{4}$

(c) 9

(d) $\frac{1}{8}$

Answer 1

Given : , $\frac{3^p}{3^n}=81$ and
To find :

Find :
By using rational components We get

By equating rational exponents we get

Now, =$\left(a^{2m+n-p}\right).\left(a^{m-2n+2p}\right)$ we get
$$\begin{gathered} ={\mathrm{a}}^{2\mathrm{m}+\mathrm{n}-\mathrm{p}+\mathrm{m}-2\mathrm{n}+2\mathrm{p}} \\ ={\mathrm{a}}^{3\mathrm{m}-\mathrm{n}+\mathrm{p}} \\ \mathrm{Now}\mathrm{putting}\mathrm{value}\mathrm{of}\mathrm{a}=2^{\frac{1}{10}}\mathrm{we}\mathrm{get}, \\ =2^{\frac{3\mathrm{m}-\mathrm{n}+\mathrm{p}}{10}} \\ =2^{\frac{6-\mathrm{n}+\mathrm{p}}{10}} \end{gathered}$$

Also, $\frac{3^p}{3^n}=81$
$3^{p-n}=3^4$
On comparing LHS and RHS we get, p - n = 4.
Now,
= a^{3m - n + p
$$\begin{gathered} =2^{\frac{6+(p-n)}{10}} \\ =2^{\frac{6+4}{10}} \\ =2^{\frac{10}{10}}=2^1 \\ =2 \end{gathered}$$}

So, option (a) is the correct answer.