Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you evaluate #log_6 (2)#?

1 Answer

A. S. Adikesavan

May 20, 2016

#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.

Related questions

See all questions in Logarithm-- Inverse of an Exponential Function

Impact of this question

5575 views around the world

You can reuse this answer
Creative Commons License