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One ordered pair (a,b) satisfies the two equations ab^4 = 384 and a^2 b^5 = 4608. What is the value of a in this ordered pair?

 Apr 2, 2020

Best Answer 

 #1
avatar+663 
+3

Hello peepeepuupuu, your username is bad.

 

I'm not sure if this is even possible, but:

(1) \(ab^4=384\)

(2) \(a^2b^5=4608\)

 

(2) divided by (1) is \(\frac{a^2b^5}{ab^4}=\frac{4608}{384}\)

\(ab=12\)

 

Plugging back in:

\(b=\frac{12}{a}\)

 

\(a(\frac{12}{a}^4)=384\)

\(12(\frac{12}{a}^3)=384\)

\((\frac{12}{a})^3=32\)

 

\(32a^3=12^3\)

\(a^3=54\)

 

\(\boxed{a=3\sqrt[3]{2}}\) I'm not sure if this is correct because I've never divided two equations before.

 Apr 2, 2020
 #1
avatar+663 
+3
Best Answer

Hello peepeepuupuu, your username is bad.

 

I'm not sure if this is even possible, but:

(1) \(ab^4=384\)

(2) \(a^2b^5=4608\)

 

(2) divided by (1) is \(\frac{a^2b^5}{ab^4}=\frac{4608}{384}\)

\(ab=12\)

 

Plugging back in:

\(b=\frac{12}{a}\)

 

\(a(\frac{12}{a}^4)=384\)

\(12(\frac{12}{a}^3)=384\)

\((\frac{12}{a})^3=32\)

 

\(32a^3=12^3\)

\(a^3=54\)

 

\(\boxed{a=3\sqrt[3]{2}}\) I'm not sure if this is correct because I've never divided two equations before.

AnExtremelyLongName Apr 2, 2020
 #2
avatar+243 
0

Thanks!! It was correct though, so yay! also my sister made the username so yee

 Apr 2, 2020
 #3
avatar+663 
+1

Nice sister

 #4
avatar+104 
+2

i'm his sister! my name is actually charis but my user is matthew macdell based on a tv show character. thank you :))

matthewmacdell  Apr 2, 2020

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