Question

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

Answer 1

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`