Question

How do you solve `e^-x = 9`?

Answer 1

I got: `x=-ln(9)=-2.19722`

Explanation:

We can take the natural log of both sides:
`ln(e^-x)=ln(9)`
where the `ln` and `e` cancel and you are left with:
`-x=ln(9)`
and so:
`x=-ln(9)=-2.19722`