jmc

Let the cubic polynomial be $P(x)=a{x}^3+b{x}^2+cx+d.$ Then $$\begin{array}{rl}a+b+c+d& =P(1),\\ 8a+4b+2c+d& =P(2),\\ 27a+9b+3c+d& =P(3),\\ 64a+16b+4c+d& =P(4),\\ 125a+25b+5c+d& =P(5).\end{array}$$Subtracting the first and second equations, second and third equations, and third and fourth equations, we get $$\begin{array}{rl}7a+3b+c& =P(2)-P(1),\\ 19a+5b+c& =P(3)-P(2),\\ 37a+7b+c& =P(4)-P(3),\\ 61a+9b+c& =P(5)-P(4).\end{array}$$Again subtracting the equations in pairs, we get $$\begin{array}{rl}12a+2b& =P(3)-2P(2)+P(1),\\ 18a+2b& =P(4)-2P(3)+P(2),\\ 24a+2b& =P(5)-2P(4)+P(3).\end{array}$$Then $$\begin{array}{rl}6a& =P(4)-3P(3)+3P(2)-P(1),\\ 6a& =P(5)-3P(4)+3P(3)-P(2),\end{array}$$so $P(5)-3P(4)+3P(3)-P(2)=P(4)-3P(3)+3P(2)-P(1).$

Hence, $$\begin{array}{rl}P(5)& =4P(4)-6P(3)+4P(2)-P(1)\\ & =4\log 4-6\log 3+4\log 2-\log 1\\ & =4\log 2^2-6\log 3+4\log 2\\ & =8\log 2-6\log 3+4\log 2\\ & =12\log 2-6\log 3\\ & =6\log 4-6\log 3\\ & =6\log \frac{4}{3}.\end{array}$$Therefore, $A+B+C=6+4+3=\boxed{13}.$