Question

Solve this ${\left(\frac{4^{m+\frac{1}{4}}\times \sqrt{{2.2}^m}}{2.\sqrt{2^{-m}}}\right)}^{\frac{1}{m}}$

Answer 1

${\left(\frac{4^{m+\frac{1}{4}}\times \sqrt{{2.2}^m}}{2.\sqrt{2^{-m}}}\right)}^{\frac{1}{m}}={\left[\frac{2^{2m+\frac{1}{2}}\times \left(2^{m+1}\right)}{{2.2}^{-\frac{m}{2}}}\right]}^{\frac{1}{m}}$
$=\frac{{(2^{2m+\frac{1}{2}}\times 2^{\frac{m+1}{2}})}^{\frac{1}{m}}}{(2^{1-\frac{m}{2}}{)}^{\frac{1}{m}}}=\frac{2^{(2m+\frac{1}{2}+\frac{m}{2}+\frac{1}{2}{)}^{\frac{1}{m}}}}{2^{\frac{1}{m}-\frac{1}{2}}}$
$=\frac{2^{2+\frac{1}{2m}+\frac{1}{2}+\frac{1}{2m}}}{2^{\frac{1}{m}-\frac{1}{2}}}=\frac{2^{\frac{5}{2}+\frac{1}{m}}}{2^{\frac{1}{m}-\frac{1}{2}}}=2^{\frac{5}{2}+\frac{1}{m}-\frac{1}{m}+\frac{1}{2}}=2^3=8$