Question

# If $\frac{{2}^{m+n}}{{2}^{n-m}}=16$, $\frac{{3}^{p}}{{3}^{n}}=81$ and $a={2}^{1/10}$, then$\frac{{a}^{2m+n-p}}{\left({a}^{m-2n+2p}{\right)}^{-1}}=$ (a) 2 (b) $\frac{1}{4}$ (c) 9 (d) $\frac{1}{8}$

Solution

## 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 Also, $\frac{{3}^{p}}{{3}^{n}}=81$ ${3}^{p-n}={3}^{4}\phantom{\rule{0ex}{0ex}}$ On comparing LHS and RHS we get, p - n = 4. Now,  = a3m - n + p $={2}^{\frac{6+\left(p-n\right)}{10}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{6+4}{10}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{10}{10}}={2}^{1}\phantom{\rule{0ex}{0ex}}=2$ So, option (a) is the correct answer.   MathematicsRD Sharma (2017)Standard IX

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