Solve for x log base x of 27=-3

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

Rewrite $\log_{x} (27) = - 3$ 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 $b^{y} = x$ .

$x^{- 3} = 27$

Solve for $x$ .

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

$\frac{1}{x^{3}} = 27$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{3}, 1$

The LCM of one and any expression is the expression.

$x^{3}$

Multiply each term in $\frac{1}{x^{3}} = 27$ by $x^{3}$ to eliminate the fractions.

Multiply each term in $\frac{1}{x^{3}} = 27$ by $x^{3}$ .

$\frac{1}{x^{3}} x^{3} = 27 x^{3}$

Simplify the left side.

Cancel the common factor of $x^{3}$ .

Cancel the common factor.

$\frac{1}{x^{3}} x^{3} = 27 x^{3}$

Rewrite the expression.

$1 = 27 x^{3}$

Solve the equation.

Rewrite the equation as $27 x^{3} = 1$ .

$27 x^{3} = 1$

Subtract $1$ from both sides of the equation.

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

Factor the left side of the equation.

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

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

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

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

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 $a = 3 x$ and $b = 1$ .

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

Simplify.

Apply the product rule to $3 x$ .

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

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

$(3 x - 1) (9 x^{2} + 3 x \cdot 1 + 1^{2}) = 0$

Multiply $3$ by $1$ .

$(3 x - 1) (9 x^{2} + 3 x + 1^{2}) = 0$

One to any power is one.

$(3 x - 1) (9 x^{2} + 3 x + 1) = 0$

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

$3 x - 1 = 0$

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

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set $9 x^{2} + 3 x + 1$ equal to $0$ .

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

Solve $9 x^{2} + 3 x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 9$ , $b = 3$ , 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 $(3 x - 1) (9 x^{2} + 3 x + 1) = 0$ true.

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

Solve for x log base x of 27=-3

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