Question

How do you solve `e^ { - 7y } = 11`?

Answer 1

`y ~~ -0.34(2dp)`

Explanation:

`e^(-7y)=11`. Taking natural log on both sides we get,

`-7y ln(e)=ln(11) or -7y= ln(11) or -7y ~~ 2.397895 [ln(e)=1]`

`:. y ~~ -2.397895/7 ~~ -0.34(2dp)` [Ans]