Solve by Factoring log base 2 of 2x^3-8-2 log base 2 of x = log base 2 of x

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

Subtract $\log_{2} (x)$ from both sides of the equation.

$\log_{2} (2 x^{3} - 8) - 2 \log_{2} (x) - \log_{2} (x) = 0$

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{2 x^{3} - 8}{x}) - 2 \log_{2} (x) = 0$

Simplify each term.

Factor $2$ out of $2 x^{3} - 8$ .

Factor $2$ out of $2 x^{3}$ .

$\log_{2} (\frac{2 (x^{3}) - 8}{x}) - 2 \log_{2} (x) = 0$

Factor $2$ out of $- 8$ .

$\log_{2} (\frac{2 x^{3} + 2 \cdot - 4}{x}) - 2 \log_{2} (x) = 0$

Factor $2$ out of $2 x^{3} + 2 \cdot - 4$ .

$\log_{2} (\frac{2 (x^{3} - 4)}{x}) - 2 \log_{2} (x) = 0$

Simplify $- 2 \log_{2} (x)$ by moving $2$ inside the logarithm.

$\log_{2} (\frac{2 (x^{3} - 4)}{x}) - \log_{2} (x^{2}) = 0$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{\frac{2 (x^{3} - 4)}{x}}{x^{2}}) = 0$

Multiply the numerator by the reciprocal of the denominator.

$\log_{2} (\frac{2 (x^{3} - 4)}{x} \cdot \frac{1}{x^{2}}) = 0$

Combine.

$\log_{2} (\frac{2 (x^{3} - 4) \cdot 1}{x \cdot x^{2}}) = 0$

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

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

$\log_{2} (\frac{2 (x^{3} - 4) \cdot 1}{x^{1} x^{2}}) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\log_{2} (\frac{2 (x^{3} - 4) \cdot 1}{x^{1 + 2}}) = 0$

Add $1$ and $2$ .

$\log_{2} (\frac{2 (x^{3} - 4) \cdot 1}{x^{3}}) = 0$

Multiply $2 (x^{3} - 4)$ by $1$ .

$\log_{2} (\frac{2 (x^{3} - 4)}{x^{3}}) = 0$

Rewrite $\log_{2} (\frac{2 (x^{3} - 4)}{x^{3}}) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ $\neq$ $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$2^{0} = \frac{2 (x^{3} - 4)}{x^{3}}$

Cross multiply to remove the fraction.

$2 (x^{3} - 4) = 2^{0} (x^{3})$

Simplify $2^{0} (x^{3})$ .

Anything raised to $0$ is $1$ .

$2 (x^{3} - 4) = 1 x^{3}$

Multiply $x^{3}$ by $1$ .

$2 (x^{3} - 4) = x^{3}$

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{3}$ from both sides of the equation.

$2 (x^{3} - 4) - x^{3} = 0$

Simplify each term.

Apply the distributive property.

$2 x^{3} + 2 \cdot - 4 - x^{3} = 0$

Multiply $2$ by $- 4$ .

$2 x^{3} - 8 - x^{3} = 0$

Subtract $x^{3}$ from $2 x^{3}$ .

$x^{3} - 8 = 0$

Factor the left side of the equation.

Rewrite $8$ as $2^{3}$ .

$x^{3} - 2^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

$(x - 2) (x^{2} + x \cdot 2 + 2^{2}) = 0$

Simplify.

Move $2$ to the left of $x$ .

$(x - 2) (x^{2} + 2 x + 2^{2}) = 0$

Raise $2$ to the power of $2$ .

$(x - 2) (x^{2} + 2 x + 4) = 0$

Simplify $(x - 2) (x^{2} + 2 x + 4)$ .

Expand $(x - 2) (x^{2} + 2 x + 4)$ by multiplying each term in the first expression by each term in the second expression.

$x \cdot x^{2} + x (2 x) + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Simplify terms.

Simplify each term.

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

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

$x^{1} x^{2} + x (2 x) + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{1 + 2} + x (2 x) + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Add $1$ and $2$ .

$x^{3} + x (2 x) + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Rewrite using the commutative property of multiplication.

$x^{3} + 2 x \cdot x + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x^{3} + 2 (x \cdot x) + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Multiply $x$ by $x$ .

$x^{3} + 2 x^{2} + x \cdot 4 - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Move $4$ to the left of $x$ .

$x^{3} + 2 x^{2} + 4 \cdot x - 2 x^{2} - 2 (2 x) - 2 \cdot 4 = 0$

Multiply $2$ by $- 2$ .

$x^{3} + 2 x^{2} + 4 x - 2 x^{2} - 4 x - 2 \cdot 4 = 0$

Multiply $- 2$ by $4$ .

$x^{3} + 2 x^{2} + 4 x - 2 x^{2} - 4 x - 8 = 0$

Combine the opposite terms in $x^{3} + 2 x^{2} + 4 x - 2 x^{2} - 4 x - 8$ .

Subtract $2 x^{2}$ from $2 x^{2}$ .

$x^{3} + 4 x + 0 - 4 x - 8 = 0$

Add $x^{3} + 4 x$ and $0$ .

$x^{3} + 4 x - 4 x - 8 = 0$

Subtract $4 x$ from $4 x$ .

$x^{3} + 0 - 8 = 0$

Add $x^{3}$ and $0$ .

$x^{3} - 8 = 0$

Add $8$ to both sides of the equation.

$x^{3} = 8$

Subtract $8$ from both sides of the equation.

$x^{3} - 8 = 0$

Factor the left side of the equation.

Rewrite $8$ as $2^{3}$ .

$x^{3} - 2^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

$(x - 2) (x^{2} + x \cdot 2 + 2^{2}) = 0$

Simplify.

Move $2$ to the left of $x$ .

$(x - 2) (x^{2} + 2 x + 2^{2}) = 0$

Raise $2$ to the power of $2$ .

$(x - 2) (x^{2} + 2 x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

$x^{2} + 2 x + 4 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x^{2} + 2 x + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x + 4$ equal to $0$ .

$x^{2} + 2 x + 4 = 0$

Solve $x^{2} + 2 x + 4 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Change the $\pm$ to $+$ .

$x = - 1 + i \sqrt{3}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Change the $\pm$ to $-$ .

$x = - 1 - i \sqrt{3}$

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

The final solution is all the values that make $(x - 2) (x^{2} + 2 x + 4) = 0$ true.

$x = 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Solve by Factoring log base 2 of 2x^3-8-2 log base 2 of x = log base 2 of x

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