# Solve the equation 1/2^(x^2-1)=square root(16^x).

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### 2 Answers

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.**

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}.**