How do you solve #22=44(1-e^(2x))#?

1 Answer
Dec 19, 2016

Answer:

I got: #x=(ln(1/2))/2=-0.34657#

Explanation:

Let us rearrange it as:
#1-e^(2x)=22/44#
#1-e^(2x)=1/2#
#e^(2x)=1-1/2#
#e^(2x)=1/2#
take the natural log of both sides:
#ln(e^(2x))=ln(1/2)#
#2xln(e)=ln(1/2)#
but #ln(e)=1#
so:
#2x=ln(1/2)#
#x=(ln(1/2))/2=-0.34657#