Question

How do you simplify `ln 1/2`?

Answer 1

I found: `-ln(2)=-0.69315` when the original question stated `ln(1/2)`...!

Explanation:

I would use a property of the logs where you have:
`logx-logy=log(x/y)`
To write:
`ln(1/2)=ln(1)-ln(2)=0-ln(2)=-ln(2)=-0.69315`

Answer 2

`0`

Explanation:

Assuming you meant `ln1/2` and not `ln(1/2)`:

`ln1=0`, so

`ln1/2=0/2=0`