How do you simplify #log_5 (1/250)#?

1 Answer
Gió
Apr 28, 2016

I found: #-log_5(2)-3#

Explanation:

We can write it as:
#log_5(1/250)=log_5(250)^-1=-log_5(250)=#
using the fact that: #log(xy)=logx+logy#
#=-log(2*125)=-[log_5(2)+log_5(125)]=#
using the definition of log we get:
#=-log_5(2)-3#

or alternatively we can use the fact that:

#log(x/y)=log(x)-log(y)#
and again:
#log(xy)=logx+logy#
and write:
#log_5(1/250)=log_5(1)-log_5(250)=#
#=log_5(1)-log_5(2*125)=#
#=log_5(1)-[log_5(2)+log_5(125)]=#
using the definition of log again we get:
#=0-log_5(2)-3=#
#=-log_5(2)-3#

Related questions
See all questions in Common Logs
Impact of this question
6055 views around the world
You can reuse this answer
Creative Commons License