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

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!