Question

What is value of `ln(e^(0.001t-3))`?

Answer 1

`0.001t-3`

Explanation:

Doing `ln(e^(f(t)))` or `e^(ln(f(t)))` always gives `f(t)`, the `ln` and `e` cancel out.

Therefore we can do: `cancel(ln)(cancel(e)^(0.001t-3)))=0.001t-3`