Question

How do you simplify `Ln(-e) - ln(-1/e)`?

Answer 1

This expression is not valid because `ln` function is only defined for positive numbers.

Explanation:

This expression is not valid because `ln` is not defined for negative numbers but if it was:

`ln(e)-ln(1/e)`

then you could use a formula: `lna-lnb=ln(a/b)` which would lead to:

`ln(e)-ln(1/e)=ln(e/(1/e))=ln(e*e)=ln(e^2)=2*lne=2`

In the last transformation I used an identity: `ln(a^b)=b*lna`

Answer 2

`ln(-e)-ln(-1/e)=2`

Explanation:

In order to simplify this, we need to use the difference law:

`loga-logb-=log(a/b)`

`ln(-e)-ln(-1/e)=ln((-e)/(-1/e))=ln(e^2)=2lne=2`

We could also use the exponent and addition laws:

`bloga-=log(a^b)` and `loga+logb-=log(ab)`

`ln(-e)-ln(-1/e)=ln(-e)+ln((-1/e)^-1)`
`=ln(-e)+ln(-e)=ln(e^2)=2`

Note that taking the `log` of negative numbers is only valid over the complex field.

Answer 3

If we are working in `CC` and using principal values of `ln(z)`, then it simplifies to `2`

Explanation:

`ln(-e) = 1+pii`

`ln(-1/e) = -1+pii`

`ln(-e) - ln(-1/e) = (1+pii)- ( -1+pii) = 2+0i = 2`