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3log ABC-logB^3 +logA-logC = log[(ABC)^3]-log(B^3)+log(A)-log(C)
= log[(A^3)(B^3)(C^3)A / (B^3)C]
= log[(A^4)(C^2)]
= 2log[(A^2)C]

3In2A-2InA
= In[(2A)^3]-In(A^2)
= In[8(A^3) / (A^2)]
= In(8A)

2007-11-28 08:04:33 · answer #1 · answered by habisce 6 · 4 1

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