How do you solve $e^{2 x} - 3 e^{x} - 4 = 0$ $e^(2x)-3e^x-4=0$ ?

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Answer 1 · 2016-12-31T11:09:12

The answer is $x = {ln 4}$ $x={ln4}$

Let $e^{x} = y$ $e^x=y$

Then the equation

$e^{2 x} - 3 e^{x} - 4 = 0$ $e^(2x)-3e^x-4=0$

becomes

$y^{2} - 3 y - 4 = 0$ $y^2-3y-4=0$

We solve this as a quadratic equation

We calculate the discriminant

$Delta=b^2-4ac=9-4*1*-4=9+16=25$

$Delta>0$ , we have 2 real solutions

$y_1=(-b+sqrtDelta)/2a=(3+5)/2=4$

$y_1=(-b-sqrtDelta)/2a=(3-5)/2=-1$

Therefore

$e^x=4$ , $=>$ , $x=ln4$

$e^x=-1$ , $=>$ , no solution as $AA x in RR, e^x>0$

How do you solve e^(2x)-3e^x-4=0e2x−3ex−4=0e^(2x)-3e^x-4=0?