Question 39231
{{{ e^2x = e^x + 2 }}}
{{{ e^2x - e^x - 2 = 0 }}}


Let {{{ y=e^x}}}, then
{{{ y^2 - y - 2 = 0 }}}
(y-2)(y+1) = 0
so y-2=0 or y+1=0
so y = 2 or y = -1


--> {{{ e^x = 2 }}} or {{{ e^x = -1 }}}
--> x = ln(2) or x = ln(-1)


Put ln(-1) into your calculator...you get an error. So the only answer is x=ln(2).


You need to plot the 2 curves on the one graph. Doing so, you will get that they cross at one point only, the solution point, at a value of x=ln(2).


jon.