Respuesta :

Beanie4185
You can essentially treat this equation as a factorable quadratic, with the variable being [tex]e^x[/tex]:

[tex]e^{2x} - 3e^x + 2 = 0[/tex]
[tex](e^x - 2)(e^x - 1) = 0[/tex]
[tex]e^x = 1, 2[/tex]
[tex]x = \ln 1, \ln 2 = 0, \ln 2[/tex]