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# help me to solve this please

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$${\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{8}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}}{{{\mathtt{16}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}}}$$

Guest Feb 28, 2015

#1
+93683
+5

Mmm

Lets see.

$$\\\dfrac{4\times 2^{1/4}\times8^{1/6}}{16^{3/4}}\\\\\\ =\dfrac{2^2\times 2^{1/4}\times(2^3)^{1/6}}{(2^4)^{3/4}}\\\\\\ =\dfrac{2^2\times 2^{1/4}\times 2^{3/6}}{2^{12/4}}\\\\\\ =\dfrac{2^2\times 2^{0.25}\times 2^{0.5}}{2^{3}}\\\\ =2^{2+0.25+0.5-3}\\\\ =2^{-0.25}\\\\ =\frac{1}{2^{1/4}}\\\\ =\frac{1}{\sqrt[4]{2}}$$

Melody  Feb 28, 2015
#1
+93683
+5

Mmm

Lets see.

$$\\\dfrac{4\times 2^{1/4}\times8^{1/6}}{16^{3/4}}\\\\\\ =\dfrac{2^2\times 2^{1/4}\times(2^3)^{1/6}}{(2^4)^{3/4}}\\\\\\ =\dfrac{2^2\times 2^{1/4}\times 2^{3/6}}{2^{12/4}}\\\\\\ =\dfrac{2^2\times 2^{0.25}\times 2^{0.5}}{2^{3}}\\\\ =2^{2+0.25+0.5-3}\\\\ =2^{-0.25}\\\\ =\frac{1}{2^{1/4}}\\\\ =\frac{1}{\sqrt[4]{2}}$$

Melody  Feb 28, 2015