MHB Help with Logarithms: Find 2^A

[IHateFactorial]

So, there's this problem:

$$A = \frac{1}{6}((\log_{2}\left({3})\right)^3 - (\log_{2}\left({6})\right)^3 - (\log_{2}\left({12})\right)^3 - (\log_{2}\left({24})\right)^3)$$

Find $$2^A$$

What I've figured out is that all the logs factorize to 3 + 2 to some power of n.

$$\log_{2}\left({3}\right)=\log_{2}\left({3\cdot2^0}\right)$$
$$\log_{2}\left({6}\right)=\log_{2}\left({3\cdot2^1}\right)$$
$$\log_{2}\left({12}\right)=\log_{2}\left({3\cdot2^2}\right)$$
$$\log_{2}\left({24}\right)=\log_{2}\left({3\cdot2^3}\right)$$

Solving THAT gives me:

$$A = \frac{1}{6}((\log_{2}\left({3})\right)^3 - (\log_{2}\left({3})+1\right)^3 - (\log_{2}\left({3})+2\right)^3 - (\log_{2}\left({3})+3\right)^3)$$

$$A = \frac{1}{6}(-2(\log_{2}\left({3})\right)^3 - 6(\log_{2}\left({3})\right)^2 - 14(\log_{2}\left({3})\right)-36$$

$$A = \frac{-1}{3}(\log_{2}\left({3})\right)^3 - (\log_{2}\left({3})\right)^2 - \frac{7}{3}(\log_{2}\left({3})\right)-6$$

And... Now? I believe I screwed up SOMEWHERE aloing that line... And I'm completely stuck as to how to go on, any ideas?

 

[Greg]

Wolfram|Alpha: Computational Knowledge Engine)

 

[anemone]

So, there's this problem:

$$A = \frac{1}{6}((\log_{2}\left({3})\right)^3 - (\log_{2}\left({6})\right)^3 - (\log_{2}\left({12})\right)^3 - (\log_{2}\left({24})\right)^3)$$

Find $$2^A$$

I am very curious and wanted to ask if you're certain that you've typed everything correctly for the expression for $A$?