Question

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

पर्याय

• 2

• \[\frac{1}{4}\]

• 9

• \[\frac{1}{8}\]

Answer 1

Given :  `2^(m+n)/2^(n-m) = 16`

\[\frac{3^p}{3^n} = 81\] and  and  `a-2^(1/10)`
To find :   `(a^(2m+n-p))/((a^(m-2n+2p))^-1)`

Find :  `2^(m+n)/2^(n-m) = 16`

By using rational components  `a^m/a^n = a^(m-n)`We get

`2^(m+n-n+m) = 16`

`2^(m+n-n+m) = 16`

`2^(2m) = 2^4`

By equating rational exponents we get 

`2m = 4`

`m = 4/2`

`m=2`

Now,  `(a(2m+n-p))/((a^(m-2n+2p))^-1`

\[\left( a^{2m + n - p} \right) . \left( a^{m - 2n + 2p} \right)\] we get

\[= a^{2m + n - p + m - 2n + 2p} \]

\[ = a^{3m - n + p} \]

\[\text { Now putting value of a } = 2^\frac{1}{10}\text { we get,} \]

\[ = 2^\frac{3m - n + p}{10} \]

\[ = 2^\frac{6 - n + p}{10}\]

Also, 

\[\frac{3^p}{3^n} = 81\]

\[3^{p - n} = 3^4 \]

On comparing LHS and RHS we get,p - n = 4.
Now,
   `(a^(2m+n-p))/(a^(m-2n+2p))^-1`= a^{3m - n + p}

\[= 2^\frac{6 + (p - n)}{10} \]

\[ = 2^\frac{6 + 4}{10} \]

\[ = 2^\frac{10}{10} = 2^1 \]

\[ = 2\]

So, option (a) is the correct answer.