If `(2^(m+n))/(2^(m-n))=16 and a=2^(1/10)` then `((a^(2m+n-p))^2)/((a^(m-2n+2p)- 1)^-1)=`

Explanation

Solution
`(2^(m+n))/(2^(n-m))=16`
We know that,
`(a^m)/(a^n)=a^(m-n)`
`(2^(m+n))/(2^(n-m))=2^(2m)`
`2^(2m)=16=2^4`
When bases are equal , power are equal
So, `2m=4`
`m=4/2`
`m=2`
`n=2`
`((a^(2m+n-p))^2)/((a^(m-2n+2p)- 1)^-1)=(a^(2m+n-p)xx(a^(2m+n-p)xx(a^(m-2n+2p)- 1)`
`=a^(2m+n-P+2m+n-P+m-2n+2P)`
`=a^(5m)`
Now, as we have `a=2^(1/10)`
We have, `2^((1/10))^(5m)`
`=2^(m/2)`
as `m=2` so, `2^(2/2)=2`.