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4^(3/2)*16^-(3/4)

Guest Jun 15, 2015

Best Answer 

 #1
avatar+14536 
+8

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

 

$${{\mathtt{4}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)} = {{\left({{\mathtt{2}}}^{{\mathtt{2}}}\right)}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}$$$${{\mathtt{2}}}^{{\mathtt{3}}}$$

 

$${{\mathtt{16}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} = {\frac{{\mathtt{1}}}{\left({\left({{\mathtt{2}}}^{{\mathtt{4}}}\right)}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}\right)}}$$      = $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}$$

 

 

$${\frac{{{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}} = {\mathtt{1}}$$

radix  Jun 15, 2015
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2+0 Answers

 #1
avatar+14536 
+8
Best Answer

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

 

$${{\mathtt{4}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)} = {{\left({{\mathtt{2}}}^{{\mathtt{2}}}\right)}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}$$$${{\mathtt{2}}}^{{\mathtt{3}}}$$

 

$${{\mathtt{16}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)} = {\frac{{\mathtt{1}}}{\left({\left({{\mathtt{2}}}^{{\mathtt{4}}}\right)}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}\right)}}$$      = $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}}$$

 

 

$${\frac{{{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{3}}}}} = {\mathtt{1}}$$

radix  Jun 15, 2015
 #2
avatar+90988 
+5

$$\\4^{3/2}*16^{-3/4}\\\\
=(4^{1/2})^3*(16^{1/4})^{-3}\\\\
=(2)^3*(2)^{-3}\\\\
=2^{3-3}\\\\
=2^0\\\\
=1$$

Melody  Jun 15, 2015

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