Solve for x 2^(1-x^2)=1/8

$2^{1 - x^{2}} = \frac{1}{8}$

Step 1

Raise $8$ to the power of $1$ .

$2^{1 - x^{2}} = \frac{1}{8^{1}}$

Step 2

Move $8^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{1 - x^{2}} = 8^{- 1}$

Step 3

Create equivalent expressions in the equation that all have equal bases.

$2^{1 - x^{2}} = 2^{- 3}$

Step 4

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$1 - x^{2} = - 3$

Step 5

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$- x^{2} = - 3 - 1$

Subtract $1$ from $- 3$ .

$- x^{2} = - 4$

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$