Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

Alice is thinking of a positive real number $x$ , and Bob is thinking of a positive real number $y$ . Given that $x^{\sqrt{y}}=27$ and $(\sqrt{x})^{y}=9$ , compute $x y$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that $27^{\sqrt{y}}=\left(x^{\sqrt{y}}\right)^{\sqrt{y}}=x^{y}=(\sqrt{x})^{2 y}=81$ so $\sqrt{y}=4 / 3$ or $y=16 / 9$ . It follows that $x^{4 / 3}=27$ or $x=9 \sqrt[4]{3}$ . The final answer is $9 \sqrt[4]{3} \cdot 16 / 9=16 \sqrt[4]{3}$ .

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