Consider #e^x# as t and the given equation is transformed into a quadratic equation #t^2# -3t+1=0. Use quadratic formula to solve for t= #(3+-sqrt5)/2#. Now substitute #e^x# for t #e^x# =#(3+-sqrt5)/2#
and take log on both sides, to get x= #ln{(3+-sqrt5)/2}#