Solve for x 8^(x^2-16)=1

$8^{x^{2} - 16} = 1$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (8^{x^{2} - 16}) = \ln (1)$

Step 2

Expand $\ln (8^{x^{2} - 16})$ by moving $x^{2} - 16$ outside the logarithm.

$(x^{2} - 16) \ln (8) = \ln (1)$

Step 3

Simplify the left side.

Apply the distributive property.

$x^{2} \ln (8) - 16 \ln (8) = \ln (1)$

Step 4

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$x^{2} \ln (8) - 16 \ln (8) = 0$

Step 5

Add $16 \ln (8)$ to both sides of the equation.

$x^{2} \ln (8) = 16 \ln (8)$

Step 6

Divide each term in $x^{2} \ln (8) = 16 \ln (8)$ by $\ln (8)$ and simplify.

Divide each term in $x^{2} \ln (8) = 16 \ln (8)$ by $\ln (8)$ .

$\frac{x^{2} \ln (8)}{\ln (8)} = \frac{16 \ln (8)}{\ln (8)}$

Simplify the left side.

Cancel the common factor of $\ln (8)$ .

Cancel the common factor.

$\frac{x^{2} \ln (8)}{\ln (8)} = \frac{16 \ln (8)}{\ln (8)}$

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

$x^{2} = \frac{16 \ln (8)}{\ln (8)}$

Simplify the right side.

Cancel the common factor of $\ln (8)$ .

Cancel the common factor.

$x^{2} = \frac{16 \ln (8)}{\ln (8)}$

Divide $16$ by $1$ .

$x^{2} = 16$

Step 7

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

$x = \pm \sqrt{16}$

Step 8

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

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

$x = \pm 4$

Step 9

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

Solve for x 8^(x^2-16)=1

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