Question

How do you evaluate `log_6 (2)`?

Answer 1

`log 2/log 6=ln 2/ln 6=0.386853`, nearly.

Explanation:

Use `log_b a=log_c a/log_c b= log a/log b=ln a/ln b`.

Here, a = 2 and b = 6.

So, `log_6 2=log 2/log 6=ln 2/ln 6=0.386853`, nearly.