How do you approximate #log_5 30# given #log_5 2=0.4307# and #log_5 3=0.6826#?

1 Answer
mason m
Nov 29, 2016

Rewrite #30# using multiples of #2#, #3#, and #5#.

#log_5(30)=log_5(2*3*5)#

Using the rule #log_a(bc)=log_a(b)+log_a(c)#, this becomes

#log_5(30)=log_5(2)+log_5(3)+log_5(5)#

Using #log_2(5)approx0.4307#, #log_3(5)approx0.6828#, and the rule that #log_a(a)=1#:

#log_5(30)approx0.4307+0.6826+1approx2.1133#

Related questions
See all questions in Common Logs
Impact of this question
3065 views around the world
You can reuse this answer
Creative Commons License