What Is the Value of $2^Y$ for Y in Problem of the Week #128?

[anemone]

Let $Y=\dfrac{1}{6}\left((\log_2 3)^3-(\log_2 6)^3-(\log_2 12)^3+(\log_2 24)^3\right)$.

Compute $2^Y$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. laura123
3. lfdahl
4. mente oscura
5. kaliprasad

Honorable mention goes also to soroban as he computed only the value of $Y$ but not $2^Y$.

Solution from MarkFL:

Spoiler

Let us first consider:

\(\displaystyle \left(\log_2\left(2^n\cdot3\right)\right)^3=\left(n+\log_2(3)\right)^3=n^3+3n^2\log_2(3)+3n\log_2^2(3)+\log_2^3(3)\)

And so we find:

\(\displaystyle \left(\log_2(3)\right)^3-\left(\log_2(6)\right)^3-\left(\log_2(12)\right)^3+\left(\log_2(24)\right)^3=\)

\(\displaystyle \log_2^3(3)-\left(1+3\log_2(3)+3\log_2^2(3)+\log_2^3(3)\right)-\left(8+12\log_2(3)+6\log_2^2(3)+\log_2^3(3)\right)+\left(27+27\log_2(3)+9\log_2^2(3)+\log_2^3(3)\right)=\)

\(\displaystyle 18+12\log_2(3)=6\left(3+2\log_2(3)\right)\)

Hence:

\(\displaystyle Y=3+2\log_2(3)=\log_2(72)\)

And thus:

\(\displaystyle 2^Y=72\)

Solution from mente oscura:

Spoiler

$$Y=\dfrac{1}{6}\left((\log_2 3)^3-(\log_2 6)^3-(\log_2 12)^3+(\log_2 24)^3\right)$$

$$Let \ a=\log_2 3$$

$$Y=\dfrac{1}{6}\left(a^3-(\log_2 2+a)^3-(\log_2 4+a)^3+(\log_2 8+a)^3\right)$$

$$Y=\dfrac{1}{6} \ (a^3-1-3a-3a^2-a^3-8-12a-6a^2-a^3+27+27a+9a^2+a^3)$$

$$Y=\dfrac{1}{6} \ (18+12a)=3+2a$$

$$Y=3+2a=3+2 \log_2 3=\log_2 8+\log_2 9=\log_2 72$$

Therefore:

$$2^Y=72$$