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}
+128472 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)
