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One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 48$ and $a^2 = 24$. What is the value of $b$ in this ordered pair?

 Mar 23, 2024

Best Answer 

 #2
avatar+128794 
+1

a = 48 / b^4

a^2  = 48^2 / b^8

 

     a^2  =  a^2

48^2 / b^8  = 24

b^8  = 48^2 / 24

b  = ± [ 48^2 / 24 ] ^(1/8)

b = ± [96]^(1/8)

 

cool cool cool

 Mar 23, 2024
 #1
avatar+16 
+1

Consider                                                                ab4  =  48  

 

We will divide both sides by ab.  

 

Since ab=72, we will divide the left side  

by "ab" and the right side by its equal 72.  

                                                                              ab4         48  

                                                                             ——   =   ——  

                                                                              ab           72  

Note that ab4 = (ab) * (b3)  

 

Cancel ab out of the left side.  

Reduce 48/72 on the right side.  

                                                                               b3           2  

                                                                             ——   =   ——  

                                                                                1            3  

 

 

                                                                                 b   =   cube root of (2 / 3)  

 Mar 23, 2024
 #2
avatar+128794 
+1
Best Answer

a = 48 / b^4

a^2  = 48^2 / b^8

 

     a^2  =  a^2

48^2 / b^8  = 24

b^8  = 48^2 / 24

b  = ± [ 48^2 / 24 ] ^(1/8)

b = ± [96]^(1/8)

 

cool cool cool

CPhill Mar 23, 2024

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