Question

How do you solve `e^ { - 4x } + 8= 35`?

Answer 1

You have to use logarithms.

Explanation:

`e^(-4x) + 8 = 35`

`e^(-4x) = 27`

`ln 27 = - 4x`

`(ln 27) / (-4) = x`

`"*"ln` is basically `log_e`

`x` is approximately `-0.8240`.