How do you solve $e^{2 x} - 4 e^{x} - 5 = 0$ $e^(2x)-4e^x-5=0$ ?

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Answer 1 · 2017-02-14T20:33:11

See explanation.

First step is to substitute $e^{x}$ $e^x$ with a new variable:

$t = e^{x}$ $t=e^x$

This makes the equation a quadratic one: $t^{2} - 4 t - 5 = 0$ $t^2-4t-5=0$

Since $e^{x}$ $e^x$ is never negative we have to choose only positive solutions of the equation.

$Delta=(-4)^2-4*1*(-5)=16+2=36$

$sqrt(Delta)=6$

$t_1=(4-6)/2=-1$

We cannot take this solution because (as I wrote earlier) $e^x$ is never negative.

$t_2=(4+6)/2=10/2=5$

This solution is a valid one, so we can now calculate the oroginal variable $x$ :

$e^x=5 =>x=ln5$

Answer: This equation has one solution $x=ln5$

How do you solve e^(2x)-4e^x-5=0e2x−4ex−5=0e^(2x)-4e^x-5=0?