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Simplify each expression.

(8b^(2/3)*9t^(1/5)) (8b^(5/3)*9t^(3/5))

 

 

 

Note: The fractions within the parenthesis simply mean they are fractions. It just looks strange within another pair of parenthesis, but I'm sure most of you have deduced that already.

 

Many thanks!!

 Mar 30, 2015

Best Answer 

 #2
avatar+1904 
+5

$$\left(\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}\right){\mathtt{\,\times\,}}\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)}\right)\right)$$

 

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)}$$

 

$${\mathtt{5\,184}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{7}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right)}$$ or $${\mathtt{5\,184}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{{\mathtt{b}}}^{{\mathtt{7}}}}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{{\mathtt{t}}}^{{\mathtt{4}}}}}$$

.
 Mar 30, 2015
 #1
avatar+128475 
+5

(8b^(2/3)*9t^(1/5)) (8b^(5/3)*9t^(3/5)) ......  rearrange.......

[8b^(2/3) * 8b^(5/3)] * [ 9t^(1/5) * 9t^(3/5) ] =

[64 b^(7/3)] * [ 81 t^(4/5) ] =

5184* b^(7/3)* t^(4/5)

And that's it.....!!!!

 

  

 Mar 30, 2015
 #2
avatar+1904 
+5
Best Answer

$$\left(\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}\right){\mathtt{\,\times\,}}\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)}\right)\right)$$

 

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)}$$

 

$${\mathtt{5\,184}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{\left({\frac{{\mathtt{7}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right)}$$ or $${\mathtt{5\,184}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{{\mathtt{b}}}^{{\mathtt{7}}}}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{{\mathtt{t}}}^{{\mathtt{4}}}}}$$

gibsonj338 Mar 30, 2015

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