Question

How do you solve `-1-8ln(-8b-5)=7`?

Answer 1

`b = 1/8(1/e + 5)`

Explanation:

`-1-8ln(-8b-5)=7`

`ln(-8b-5)= -1`

`"Note" : log_b a = c => a = b^c`

`-8b-5 = e^-1`

`b = 1/8(1/e + 5)`