Question 353796
The equation is actually a quadratic equation in {{{3^x}}}.  It is factorable.
{{{3^(2x)+3^x-2=0}}},
{{{(3^x+2)*(3^x-1)=0}}}, 
This means {{{3^x + 2=0}}}, or {{{3^x - 1=0}}}.  The first equation has no solution, because {{{3^x + 2>0}}}, while the second gives
{{{3^x = 1}}}.  Only {{{x=0}}} will satisfy this equation.