Solve for X 3 log base 3 of X=- log base 3 of 27

$3 \log_{3} (X) = - \log_{3} (27)$

Simplify.

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

$\log_{3} (X^{3}) = - \log_{3} (27)$

Simplify $- \log_{3} (27)$ by moving $- 1$ inside the logarithm.

$\log_{3} (X^{3}) = \log_{3} (27^{- 1})$

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

$X^{3} = 27^{- 1}$

Solve for $X$ .

Subtract $27^{- 1}$ from both sides of the equation.

$X^{3} - 27^{- 1} = 0$

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

$X^{3} - \frac{1}{27} = 0$

Factor the left side of the equation.

Rewrite $\frac{1}{27}$ as ${(\frac{1}{3})}^{3}$ .

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

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 $b = \frac{1}{3}$ .

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

Simplify.

Combine $X$ and $\frac{1}{3}$ .

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

Apply the product rule to $\frac{1}{3}$ .

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

One to any power is one.

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

Raise $3$ to the power of $2$ .

$(X - \frac{1}{3}) (X^{2} + \frac{X}{3} + \frac{1}{9}) = 0$

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

$X - \frac{1}{3} = 0$

$X^{2} + \frac{X}{3} + \frac{1}{9} = 0$

Set $X - \frac{1}{3}$ equal to $0$ and solve for $X$ .

Set $X - \frac{1}{3}$ equal to $0$ .

$X - \frac{1}{3} = 0$

Add $\frac{1}{3}$ to both sides of the equation.

$X = \frac{1}{3}$

Set $X^{2} + \frac{X}{3} + \frac{1}{9}$ equal to $0$ and solve for $X$ .

Set $X^{2} + \frac{X}{3} + \frac{1}{9}$ equal to $0$ .

$X^{2} + \frac{X}{3} + \frac{1}{9} = 0$

Solve $X^{2} + \frac{X}{3} + \frac{1}{9} = 0$ for $X$ .

Multiply through by the least common denominator $9$ , then simplify.

Apply the distributive property.

$9 X^{2} + 9 (\frac{X}{3}) + 9 (\frac{1}{9}) = 0$

Simplify.

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$9 X^{2} + 3 (3) (\frac{X}{3}) + 9 (\frac{1}{9}) = 0$

Cancel the common factor.

$9 X^{2} + 3 \cdot (3 (\frac{X}{3})) + 9 (\frac{1}{9}) = 0$

Rewrite the expression.

$9 X^{2} + 3 X + 9 (\frac{1}{9}) = 0$

Cancel the common factor of $9$ .

Cancel the common factor.

$9 X^{2} + 3 X + 9 (\frac{1}{9}) = 0$

Rewrite the expression.

$9 X^{2} + 3 X + 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 9$ , $b = 3$ , and $c = 1$ into the quadratic formula and solve for $X$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (9 \cdot 1)}}{2 \cdot 9}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$X = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}$

Multiply $- 4$ by $9$ .

$X = \frac{- 3 \pm \sqrt{9 - 36 \cdot 1}}{2 \cdot 9}$

Multiply $- 36$ by $1$ .

$X = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 9}$

Subtract $36$ from $9$ .

$X = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 9}$

Rewrite $- 27$ as $- 1 (27)$ .

$X = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 9}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$X = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 9}$

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

$X = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 9}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$X = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 9}$

Rewrite $9$ as $3^{2}$ .

$X = \frac{- 3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 9}$

Pull terms out from under the radical.

$X = \frac{- 3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 9}$

Move $3$ to the left of $i$ .

$X = \frac{- 3 \pm 3 i \sqrt{3}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$X = \frac{- 3 \pm 3 i \sqrt{3}}{18}$

Simplify $\frac{- 3 \pm 3 i \sqrt{3}}{18}$ .

$X = \frac{- 1 \pm i \sqrt{3}}{6}$

The final answer is the combination of both solutions.

$X = - \frac{1 - i \sqrt{3}}{6}, - \frac{1 + i \sqrt{3}}{6}$

The final solution is all the values that make $(X - \frac{1}{3}) (X^{2} + \frac{X}{3} + \frac{1}{9}) = 0$ true.

$X = \frac{1}{3}, - \frac{1 - i \sqrt{3}}{6}, - \frac{1 + i \sqrt{3}}{6}$

Solve for X 3 log base 3 of X=- log base 3 of 27

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