Solve equation e^(-x^2)=e^(-3x-4)
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.
-x^2 = -3x - 4
==> x^2 - 3x - 4 = 0
==> (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^(-16)
e^(-x2^2)=e^(-3x2-4) for x2 = -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-4)(x+1) = 0
x-4 =0 or x+1 = 0
x=4 or x =-1.