How do you solve $e^(-4x)+8=35$ ?

Question

2208 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T15:52:43

$x=-0.824$

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

$hArre^(-4x)=35-8=27$

Hence, $-4x=ln27=3.2958369$ from scientific calculator

$x=-3.2958369/4=-0.824$

How do you solve e^(-4x)+8=35e^(-4x)+8=35?