Answer to Question #3661 in Algebra for tami

Question

Solve 2e[sup]2x[/sup] + 19e[sup]x[/sup]-27=0.
Moderate step by step would be great! Thanks!

Answer 1

Denote e^x as t, thus the equation is
2t² +19t -27 = 0
D = 19² + 4*2*27 = 577
t₁ = (-19 + √577)/4
t₂ = (-19 - √577)/4 < 0 - doesnt's satisfy the conditions, as e^x >=0 at any x.

e^x = (-19 + √577)/4, x = ln(-19 + √577) - ln4