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

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

Step 1

Simplify.

Simplify $3 \log_{2} (x)$ by moving $3$ inside the logarithm.

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

Simplify $- \log_{2} (8)$ by moving $- 1$ inside the logarithm.

$\log_{2} (x^{3}) = \log_{2} (8^{- 1})$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{3} = 8^{- 1}$

Step 3

Solve for $x$ .

Subtract $8^{- 1}$ from both sides of the equation.

$x^{3} - 8^{- 1} = 0$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x^{3} - \frac{1}{8} = 0$

Factor the left side of the equation.

Rewrite $\frac{1}{8}$ as ${(\frac{1}{2})}^{3}$ .

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

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 $b = \frac{1}{2}$ .

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

Simplify.

Combine $x$ and $\frac{1}{2}$ .

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

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

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

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

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

$x - \frac{1}{2} = 0$

$x^{2} + \frac{x}{2} + \frac{1}{4} = 0$

Set $x - \frac{1}{2}$ equal to $0$ and solve for $x$ .

Set $x - \frac{1}{2}$ equal to $0$ .

$x - \frac{1}{2} = 0$

Add $\frac{1}{2}$ to both sides of the equation.

$x = \frac{1}{2}$

Set $x^{2} + \frac{x}{2} + \frac{1}{4}$ equal to $0$ and solve for $x$ .

Set $x^{2} + \frac{x}{2} + \frac{1}{4}$ equal to $0$ .

$x^{2} + \frac{x}{2} + \frac{1}{4} = 0$

Solve $x^{2} + \frac{x}{2} + \frac{1}{4} = 0$ for $x$ .

Multiply through by the least common denominator $4$ , then simplify.

Apply the distributive property.

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

Simplify.

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$4 x^{2} + 2 x + 1 = 0$

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 4 \cdot 1$ .

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 4}$

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

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

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

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

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

Factor $4$ out of $12$ .

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

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{8}$ .

$x = \frac{- 1 \pm i \sqrt{3}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

The final solution is all the values that make $(x - \frac{1}{2}) (x^{2} + \frac{x}{2} + \frac{1}{4}) = 0$ true.

$x = \frac{1}{2}, - \frac{1 - i \sqrt{3}}{4}, - \frac{1 + i \sqrt{3}}{4}$

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

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