Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

Let $A=\frac{1}{6}\left(\left(\log _{2}(3)\right)^{3}-\left(\log _{2}(6)\right)^{3}-\left(\log _{2}(12)\right)^{3}+\left(\log _{2}(24)\right)^{3}\right)$ Compute $2^{A}$ .

Your answer

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

Solution

Let $a=\log _{2}(3)$ , so $2^{a}=3$ and $A=\frac{1}{6}\left[a^{3}-(a+1)^{3}-(a+2)^{3}+(a+3)^{3}\right]$ . But $(x+1)^{3}-x^{3}=3 x^{2}+3 x+1$ , so $A=\frac{1}{6}\left[3(a+2)^{2}+3(a+2)-3 a^{2}-3 a\right]=\frac{1}{2}[4 a+4+2]=2 a+3$ . Thus $2^{A}=\left(2^{a}\right)^{2}\left(2^{3}\right)=9 \cdot 8=72$

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