An algebra problem by Pandu Christian

$e^{2x}$ + $e^{x}$ $= 6$

$(e^{x})^{2}$ + $e^{x}$ $= 6$

$e^{x}$ $= y$

$y^{2}$ $+ y = 6$

$y^{2} + y - 6 = 0$

$(y-2)(y+3) = 0$

$y-2=0$ $or$ $y+3=0$

$y =2$ $or$ $y = -3$

$e^{x} =2$ $or$ $e^{x} = -3$

$e^{x} = 2$

$x = ln(2)$