Question

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

• A. $2$
• B. $\frac{1}{4}$
• C. $9$
• D. None of the above

Answer 1

Answer: (D)

Given:

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

$\frac{3^p}{3^n}=81$,

$a=2^{\frac{1}{10}}$.

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

We know that

$\frac{a^m}{a^n}$ = $a^{m-n}$

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

$\Rightarrow 2^{2m}$ = 16 = $2^4$

When bases are equal, Powers are Equal

$\Rightarrow 2m=4$

$\Rightarrow m=2$ ---(1)

Now, lets take, $\frac{3^p}{3^n}=81$

$\Rightarrow 3^{p-n}=3^4$

$\Rightarrow p-n=4$ [Since, When bases are equal, Powers are Equal.]

$\Rightarrow n-p=-4$ ---(2)

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

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

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

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

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

$=a^{3m-n+p}$

$=a^{3m-(n-p)}$

$=a^{3\left(2\right)-(-4)}$

$=a^{10}$

$=(2^{\frac{1}{10}}{)}^{10}$ [$\because a=2^{\frac{1}{10}}$]

$=2^{\frac{1}{10}\times 10}$

$=2$

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

Hence, Option A is correct.