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}".
Lol are you part of the AoPS algebra class?
I am in intro to alg a
This is literally a homework question for AOPS
BRO WHATS THE ANSWER. I DO AOPS AND THIS QUESTION IS HECKA ANNOYING
ME TOOOOOOOOOOOOO
Raising the first equation to the fifth power, we get $a^5 b^{20} = 12^5 = 248832$. Dividing this equation by the second equation, we get $b^{15} = 32$, so $b = \sqrt[15]{32} = \sqrt[15]{2^5} = \sqrt[3]{2}$. Then from the first equation,
\[a = \frac{12}{b^4} = \frac{12}{(\sqrt[3]{2})^4} = \frac{12}{\sqrt[3]{2^4}} = \frac{12}{2 \sqrt[3]{2}} = \frac{6}{\sqrt[3]{2}}.\]We can simplify this fraction by multiplying the top and bottom by $\sqrt[3]{4}$:
\[a = \frac{6}{\sqrt[3]{2}} = \frac{6 \sqrt[3]{4}}{\sqrt[3]{2} \cdot \sqrt[3]{4}} = \frac{6 \sqrt[3]{4}}{\sqrt[3]{8}} = \frac{6 \sqrt[3]{4}}{2} = 3 \sqrt[3]{4}.\]Therefore, the solution is $(a,b) = (3 \sqrt[3]{4}, \sqrt[3]{2})$; the desired value of $a$ is $\boxed{3\sqrt[3]{4}}$.
Raising the first equation to the fifth power, we get $a^5 b^{20} = 12^5 = 248832$. Dividing this equation by the second equation, we get $b^{15} = 32$, so $b = \sqrt[15]{32} = \sqrt[15]{2^5} = \sqrt[3]{2}$. Then from the first equation,
\[a = \frac{12}{b^4} = \frac{12}{(\sqrt[3]{2})^4} = \frac{12}{\sqrt[3]{2^4}} = \frac{12}{2 \sqrt[3]{2}} = \frac{6}{\sqrt[3]{2}}.\]We can simplify this fraction by multiplying the top and bottom by $\sqrt[3]{4}$:
\[a = \frac{6}{\sqrt[3]{2}} = \frac{6 \sqrt[3]{4}}{\sqrt[3]{2} \cdot \sqrt[3]{4}} = \frac{6 \sqrt[3]{4}}{\sqrt[3]{8}} = \frac{6 \sqrt[3]{4}}{2} = 3 \sqrt[3]{4}.\]Therefore, the solution is $(a,b) = (3 \sqrt[3]{4}, \sqrt[3]{2})$; the desired value of $a$ is $\boxed{3\sqrt[3]{4}}$.