Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Find the answer

If $a$ and $b$ are positive real numbers such that $a \cdot 2^{b}=8$ and $a^{b}=2$ , compute $a^{\log _{2} a} 2^{b^{2}}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Taking $\log _{2}$ of both equations gives $\log _{2} a+b=3$ and $b \log _{2} a=1$ . We wish to find $a^{\log _{2} a} 2^{b^{2}}$ ; taking $\log _{2}$ of that gives $\left(\log _{2} a\right)^{2}+b^{2}$ , which is equal to $\left(\log _{2} a+b\right)^{2}-2 b \log _{2} a=3^{2}-2=7$ . Hence, our answer is $2^{7}=128$ .

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