Question 1038416
Use a substitution,
{{{u=e^x}}}
Then,
{{{u^2=e(2x)}}}
So,
{{{2u^2-3u+1=0}}}
{{{(u-1)(2u-1)=0}}}
Two solutions:
{{{u-1=0}}}
{{{u=1}}}
{{{e^x=1}}}
{{{x=0}}}
and
{{{2u-1=0}}}
{{{2u=1}}}
{{{u=1/2}}}
{{{e^x=1/2}}}
{{{x=ln(1/2)}}}