Solve for x log base 2 of 8^x=-3

$\log_{2} (8^{x}) = - 3$

Expand $\log_{2} (8^{x})$ .

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

$x \log_{2} (8)$

Logarithm base $2$ of $8$ is $3$ .

$x \cdot 3$

The expanded equation is $x \cdot 3 = - 3$ .

$x \cdot 3 = - 3$

Divide each term in $(x \cdot 3) = - 3$ by $3$ and simplify.

Divide each term in $(x \cdot 3) = - 3$ by $3$ .

$\frac{(x \cdot 3)}{3} = \frac{- 3}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x \cdot 3}{3} = \frac{- 3}{3}$

Divide $x$ by $1$ .

$x = \frac{- 3}{3}$

Simplify the right side.

Divide $- 3$ by $3$ .

$x = - 1$