How do you solve #e^ { 4x } - 3e ^ { 2x } - 18= 0#?

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Answer 1 · 2018-05-07T09:09:54

#color(brown)(x = (1/2) log _e 6 ~~ 0.8959#

#e^(4x) - 3 e^(2x) - 18 = 0#

Let #e^(2x) = a#

#"The equation becomes " a^2 - 3a - 18 = 0#

#a^2 + 3a - 6a - 18 = 0#

#a (a + 3) - 6 (a + 3) = 0#

(a-6) * (a + 3) = 0#

#a = 6, -3#

#a = e^(2x) = 6

#2x = log _e 6#

#color(brown)(x = (1/2) log _e 6 = 1.791759469228055/2 ~~ 0.8959#

#a = e^2x = -3#

#2x = log _e -3#

#x = (1/2) log _e -3, " imaginary"#