Question 420934
Denote 3^x=y and our equation transformed: y^2-1=Y+2 or y^2-y-3=0, which is a quadratic equation. Solving it we have: y=(1+sqrt(13))/2, y=(1-sqrt(13))/2

  Therefore:  3^x=(1+sqrt(13))/2 and 3^x=(1-sqrt(13))/2

   xln3=ln(1+sqrt(13))/2 and  xln3=ln(1-sqrt(13))/2

x=(ln(1+sqrt(13))/ln(9)  and x=(ln(1-sqrt(13))/ln(9).