Question #3661

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

Expert's answer

Denote e^{x} as t, thus the equation is

2t^{2} +19t -27 = 0

D = 19^{2} + 4*2*27 = 577

t_{1} = (-19 + √577)/4

t_{2} = (-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

2t

D = 19

t

t

e

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