# If 2 M + N 2 N − M = 16 , 3 P 3 N = 81 and a = 2 1 / 10 ,Than a 2 M + N − P ( a M − 2 N + 2 P ) − 1 = - Mathematics

MCQ

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

#### Options

• 2

• $\frac{1}{4}$
• 9

• $\frac{1}{8}$

#### Solution

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= a3m - 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.

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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 2 Exponents of Real Numbers
Q 35 | Page 32