Question

How do you solve `15e^-x=645`?

Answer 1

`x~=-3.76`

Explanation:

`e^-x=645/15=43`

We take logs of both sides:

`loge^-x=log43~=3.76`

But `loge^-x`, by definition, is the power to which we raise the base `e` to get `e^-x`, so here `-x~=3.76`.

And `x~=-3.76`