Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you evaluate #log_343 (1/7)#?

1 Answer

Nov 13, 2016

#-1/3#

Explanation:

Let #n = log_343 (1/7)#

#343^n=1/7#

Rewritten using all bases of 7:
#(7^3)^n=7^(-1)#

#7^(3n)=7^(-1)#

#3n=-1#

#n=-1/3#

Related questions

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

Impact of this question

2751 views around the world

You can reuse this answer
Creative Commons License