Question

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

• A. 2
• B. $\frac{1}{4}$
• C. 9
• D. $\frac{1}{8}$

Answer 1

$\frac{2^{m+n}}{2^{n-m}}=16,\frac{3^p}{3^n}=81anda=2^{\frac{1}{10}},then\frac{a^{2m+n-p}}{(a^{m-2n+2p}{)}^{-1}}\Rightarrow 2^{m+n-n+m}=2^4\Rightarrow 2^{2m}=2^{4}Comparing,weget2m=4\Rightarrow m=\frac{4}{2}=2Given,\frac{3^p}{3^n}=813^{p-n}=(3{)}^{4}p-n=4\frac{a^{2m+n-p}}{(a^{m-2n+2p}{)}^{-1}}=a^{2m+n-p}\times a^{m-2n+2p}=a^{2m+n-p+m-2n+2p}=a^{3m-n+p}\because a=2^{\frac{1}{10}}\Rightarrow a^{3m-n+p}={\left(2^{\frac{1}{10}}\right)}^{3m-n+p}=2^{\frac{3m-n+p}{10}}=2^{\frac{3\times 2-n+p}{10}}=2^{\frac{6-n+p}{10}}=2^{\frac{6+4}{10}}=2^{\frac{10}{10}}=2$