# Solve equation e^(-x^2)=e^(-3x-4)

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e^(-x^2) = e^(-3x-4)

We know that:

if e^a = e^b

==> a = b

Since the bases are equals, then the powers should be equal.

=Therefore,

-x^2 = -3x - 4

==> x^2 - 3x - 4 = 0

Factor:

==> (x-4)(x+1) = 0

Then we have two solutions:

==> **x1= 4 **

==> **x2= -1**

This is an exponential equation.

We notice that the bases from the both sides are matching, so we'll use the one to one property of exponential functions:

e^(-x^2)=e^(-3x-4) => -x^2 = -3x-4

We'll move all terms to one side:

-x^2+3x+4 = 0

We'll multiply by (-1):

x^2 - 3x - 4 = 0

We'll apply the quadratic formula:

x1 = [3+sqrt(9+16)]/2

x1 = (3+5)/2

x1 = 4

x2 = (3-5)/2

x2 = -1

We'll check the solutions into the original equation:

e^(-x1^2)=e^(-3x1-4) for **x1 = 4**

e^(-16)=e^(-12-4)

e^(-16) = e^(-16)

e^(-x2^2)=e^(-3x2-4) for **x2 = -1**

e^(-1)=e^(3-4)

e^(-1)=e^(-1)

**So, both solution are admissible!**

e^(-x^2) = e^(-3x-4). To solve for x.

Both sides of the equation are the exponent of the same base. So the exponents -x^2 on left and the exponent -3x-4 should be equal for for the equality.

-x^2 = -3x-4. Multiply by -1.

x^2 =3x+4

x^2-3x-4 =0

(x-4)(x+1) = 0

x-4 =0 or x+1 = 0

x=4 or x =-1.