Seja \(a \ \epsilon \mathbb{R}\), com a>1. Se b=log2a , então o valor de: log4a3 + log24a + log2(a/(a+1)) + (log8a)2 -log1/2((a2-1)/(a-1)) vale:
2b-3
\(\frac{65b}{18}+2\)
\(\frac{2b^2 -3b+1}{2}\)
\(\frac{2b^2 +63b +36}{18}\)
\(\frac{b^2 +9b +7}{9}\)