How do you calculate #log_6 (2)#?

1 Answer
Apr 29, 2016


log 2/log 6 = 0.386853, nearly


Use #log_b a = log_c a/log_c b#, where c is at your choice.

Let us choose c as either 10 or e. Then

#log_6 2=log_10 2/log_10 6=ln 2/ln 6=0.386853#, nearly.

I am sure that this comparison, using different bases, would reinforce the formula used, in the long term memory.