Calcular cologaritmos espero su apoyo

$$\begin{align}&A.\\&colog_3 (1/27) =-log_3 (1/27) = 3  \\&Resolviendo\\&-log_3 (1/27) \\&-(-log_3 27) -------> 3^3=27\\&log_3 (3^3) \\&3log_3 (3)\\&3*1=3\\&Entonces\\&-log_3 (1/27) = 3  \\&\\&B.\\&Colog_3 5 + log_3 5\\&-log_3 5 + log_3 5 = 0\\&\\&C.\\&colog_2 (4)^-3 = -log_2 (4)^-3 = 6\\&\\&-log_2 (4)^-3\\&-\left(-3\log _2\left(4\right)\right)\\&3\log _2\left(4\right)\\&Simlificamos\\&\log _2\left(4\right)\\&\log _2\left(2^2\right)\\&2\log _2\left(2\right)\\&2\cdot \:1 = 2\\&\\&\\&\end{align}$$

Recordemos que el 3 de la expresión no ha desaparecido, entonces:

3*2= 6