Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve #log_7n=2/3log_7 8#?

1 Answer

Nov 24, 2016

#n =4#

Explanation:

#log_7 n = 2/3log_7 8" "larr# use the log power law

#log_7 n = log_7 8^(2/3)#

#:. n = 8^(2/3)" "larr logA = logB hArr A = B#

#n = (root3 8)^2#

#n = 2^2#

#n = 4#

Related questions

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

Impact of this question

3272 views around the world

You can reuse this answer
Creative Commons License