How do you solve for x?: #e^(-2x)+e^x = 1/3?#

1 Answer
Oct 22, 2015

Answer:

I found: #-ln[3(e^-2+1)]=-ln(3)-ln(e^-2+1)#

Explanation:

Collect #e^x# on the left:
#e^x(e^-2+1)=1/3# rearrange:
#e^x=1/(3(e^-2+1))#
apply the #ln# on both sides:
#ln[e^x]=ln[1/(3(e^-2+1))]#
so:
#x=ln[1/(3(e^-2+1))]#
use the properties of log relating sums and subtractions to multiplications and divisions to get:
#x=ln(1)-ln[3(e^-2+1)]=0-ln(3)-ln(e^-2+1)#