One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 48$ and $a^2 b^5 = 72$. What is the value of $b$ in this ordered pair?
We can divide the two equations to get:
a^2/b = 2
Squaring both sides, we get:
a^4/b^2 = 4
Substituting ab^4=48, we get:
48/b^2 = 4
Solving for b2, we get:
b^2 = 12
Therefore, b = sqrt(12) = 2*sqrt(3).
+170 We divide the equations to get \(ab=\frac32\), then \((ab)b^3=48 \implies b^3 = 32\), so \(b=2\sqrt[3]4\).
