Precalculus Examples

$\log ({(x)}^{3}) = 6 \log (x)$

Step 1

Simplify $6 \log (x)$ by moving $6$ inside the logarithm.

$\log (x^{3}) = \log (x^{6})$

Step 2

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

$x^{3} = x^{6}$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $x^{6}$ from both sides of the equation.

$x^{3} - x^{6} = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $x^{3}$ out of $x^{3} - x^{6}$ .

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Step 3.2.1.1

Multiply by $1$ .

$x^{3} \cdot 1 - x^{6} = 0$

Step 3.2.1.2

Factor $x^{3}$ out of $- x^{6}$ .

$x^{3} \cdot 1 + x^{3} (- x^{3}) = 0$

Step 3.2.1.3

Factor $x^{3}$ out of $x^{3} \cdot 1 + x^{3} (- x^{3})$ .

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

Step 3.2.2

Rewrite $1$ as $1^{3}$ .

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

Step 3.2.3

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 $a = 1$ and $b = x$ .

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

Step 3.2.4

Factor.

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Step 3.2.4.1

Simplify.

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Step 3.2.4.1.1

One to any power is one.

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

Step 3.2.4.1.2

Multiply $x$ by $1$ .

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

Step 3.2.4.2

Remove unnecessary parentheses.

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

Step 3.3

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

$x^{3} = 0$

$1 - x = 0$

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

Step 3.4

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Step 3.4.2

Solve $x^{3} = 0$ for $x$ .

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Step 3.4.2.1

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

$x = \sqrt[3]{0}$

Step 3.4.2.2

Simplify $\sqrt[3]{0}$ .

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Step 3.4.2.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 3.4.2.2.2

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

$x = 0$

Step 3.5

Set $1 - x$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Step 3.5.2

Solve $1 - x = 0$ for $x$ .

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Step 3.5.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 3.5.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 3.5.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

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

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 3.5.2.2.2.2

Divide $x$ by $1$ .

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

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 3.6

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

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Step 3.6.1

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

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

Step 3.6.2

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

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Step 3.6.2.1

Use the quadratic formula to find the solutions.

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

Step 3.6.2.2

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

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

Step 3.6.2.3

Simplify.

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Step 3.6.2.3.1

Simplify the numerator.

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Step 3.6.2.3.1.1

One to any power is one.

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

Step 3.6.2.3.1.2

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

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Step 3.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 3.6.2.3.1.3

Subtract $4$ from $1$ .

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

Step 3.6.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 3.6.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Step 3.6.2.3.1.6

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

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

Step 3.6.2.3.2

Multiply $2$ by $1$ .

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

Step 3.6.2.4

The final answer is the combination of both solutions.

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

Step 3.7

The final solution is all the values that make $x^{3} (1 - x) (1 + x + x^{2}) = 0$ true.

$x = 0, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$