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

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.