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?

peepeepuupuu Apr 2, 2020

#1**+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.**

AnExtremelyLongName Apr 2, 2020

#1**+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**0 **

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

peepeepuupuu Apr 2, 2020

#4**+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