How do you solve #600/(1+e^-x)=550#?

1 Answer
Nov 10, 2017

#2.398# #("to 2d.p")#

Explanation:

We start off with #600/(1+e^-x)=550#

We want the #x# on its own, and it isn't useful in a denominator, so we will switch the #550# with the #1+e^(-x)# to get:
#600/550=12/11=1+e^(-x)#

We still need the numbers on one side, so we take away 1 from each side:
#e^(-x)=12/11-1=1/11#

To remove the #e# we take the #ln# of both sides:
#ln(e^(-x))=-x=ln(1/11)#

Now, we just have to divide both sides by -1 to get:
#x=-ln(1/11)=2.398# #("to 2d.p")#