How do you evaluate $log_6 (2)$ ?

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Answer 1 · 2016-05-20T07:43:09

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

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.

How do you evaluate log_6 (2)log_6 (2)?