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One ordered pair (a,b) satisfies the two equations $ab^4 = 12$ and $a^5 b^5 = 7776$. What is the value of $a$ in this ordered pair? Note: You can enter radicals with the \sqrt command. For example, \sqrt[3]{5} gives you $\sqrt[3]{5}$. Therefore, you can enter $2\sqrt[4]{6}$ as "2\sqrt[4]{6}".

 

Its not   \sqrt[3]{4}

 May 10, 2019
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ab^4  = 12           a^5b^5  = 7776

 

Using the first equation    a =  12/ b^4         sub this into the second equation

 

(12/b^4)^5 * b^5  = 7776

 

248832 * b^5 / b^20  = 7776

 

248832 / b^15  = 7776

 

b^15 =   248832 / 7776 

 

b^15  =  32      

 

b^15  =  2^5    

 

b =  (2^5)^(1/15)   =  ∛2   

 

And a   =  12 / (2^1/3)^4      =  12 / (2^4/3)  =  12 / (2 ∛2)   =   6/∛2  =  3*(2)^(2/3)

 

So....assuming real solutions....a  = 3*(2)^(2/3)

 

 

cool cool cool

 May 10, 2019

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