Solve the equation 1/2^(x^2-1)=square root(16^x).
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to solve the equation 1/2^(x^2-1) = sqrt (16^x)
1/2^(x^2-1) = sqrt (16^x)
=> 2^-(x^2 - 1) = 16^x^(1/2)
=> 2^(-x^2 + 1) = 2^4^x^(1/2)
=> 2^(-x^2 + 1) = 2^2x
We can equate the exponent as the base is the same
-x^2 + 1 = 2x
=> x^2 + 2x - 1 = 0
x1 = [-2 + sqrt (4 + 4)]/2
=> x1 = -1 + sqrt 2
x2 = -1 - sqrt 2
The required solution is x = -1 + sqrt 2 and x = -1 - sqrt 2.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll manipulate the left side using the negative power property:
1/2^(x^2-1)=2^-(x^2-1)
Now, we'll manipulate the right side, writting 16 as a power of 2:
sqrt(16^x) = sqrt(2^4x) = (2^4x)^(1/2) = 2^2x
We'll re-write the equation:
2^-(x^2-1) = 2^2x
Since the bases are matching now, we'll apply one to one property:
-x^2+1 = 2x
x^2 + 2x - 1 = 0
x1 = [-2+sqrt(4 + 4)]/2
x1 = (-2+2sqrt2)/2
x1 = -1+sqrt2
x2 = -1-sqrt2
The solutions of exponential equation are { -1-sqrt2 ; -1+sqrt2}.