How do you expand #Log_b( sqrt57/74)#?

1 Answer
A. S. Adikesavan
May 20, 2016

#((1/2) log 57 - log 74)/log b=((1/2) ln 57 - ln 74)/ln b#

Explanation:

Use #log_b a=log_c a/log_c b and log_b a^n=n log_b a#.

Here, the given logarithm

#= log_b( sqrt57 /74)#

#=log( sqrt57 /74)/log b=ln( sqrt57 /74)/ln b#

#=( log 57/2 - log 74)/log b=( ln 57/2 - ln 74)/ln b#

Related questions
See all questions in Functions with Base b
Impact of this question
1482 views around the world
You can reuse this answer
Creative Commons License