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((64^(3/4)*(sqrt(2)/2))^(-1/2)*(4sqrt(2))^3*sqrt(2))^(1/3)

I need to solve this, but i simply don't know how to start. So anyone can solve it, and post here every step of it?

Thanks.

Guest Dec 28, 2014

Best Answer 

 #1
avatar+91049 
+13

$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

mmm that looks scary doesn't it. :)

 

$${\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}}{{{\mathtt{2}}}^{{\mathtt{1}}}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${\left({{\mathtt{2}}}^{{\mathtt{6}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{8}}}{{\mathtt{12}}}}\right)}$$

 

Now    -9+1+24+8=24

 

$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$

Melody  Dec 28, 2014
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7+0 Answers

 #1
avatar+91049 
+13
Best Answer

$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

mmm that looks scary doesn't it. :)

 

$${\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}}{{{\mathtt{2}}}^{{\mathtt{1}}}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

 

 

$${{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${\left({{\mathtt{2}}}^{{\mathtt{6}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{8}}}{{\mathtt{12}}}}\right)}$$

 

Now    -9+1+24+8=24

 

$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$

 

$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$

Melody  Dec 28, 2014
 #2
avatar+91049 
+10

By doing it so slowly to show you all the steps it actually made it seem much harder. :)

Melody  Dec 28, 2014
 #3
avatar+11753 
+10

Melody........that was a HUGE answer!

 

rosala  Dec 28, 2014
 #4
avatar+91049 
+5

thanks Rosala - it was a bit of an effort :)

Melody  Dec 28, 2014
 #5
avatar+11753 
+5

surely it might have been!

 

rosala  Dec 28, 2014
 #6
avatar+78750 
+5

Wow, Melody....!!!..that was a lot of work just to arrive at "4"......but I'll give you credit.....you even managed to keep rosala's attention....!!!  LOL!!!

 

CPhill  Dec 28, 2014
 #7
avatar+91049 
+5

Thanks Chris and rosala,

It was a pretty cool question.   

Melody  Dec 29, 2014

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