Solve for x log base x of 216=3

$\log_{x} (216) = 3$

Step 1

Rewrite $\log_{x} (216) = 3$ 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 $b^{y} = x$ .

$x^{3} = 216$

Step 2

Solve for $x$ .

Subtract $216$ from both sides of the equation.

$x^{3} - 216 = 0$

Factor the left side of the equation.

Rewrite $216$ as $6^{3}$ .

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

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 $b = 6$ .

$(x - 6) (x^{2} + x \cdot 6 + 6^{2}) = 0$

Simplify.

Move $6$ to the left of $x$ .

$(x - 6) (x^{2} + 6 x + 6^{2}) = 0$

Raise $6$ to the power of $2$ .

$(x - 6) (x^{2} + 6 x + 36) = 0$

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

$x - 6 = 0$

$x^{2} + 6 x + 36 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x^{2} + 6 x + 36$ equal to $0$ and solve for $x$ .

Set $x^{2} + 6 x + 36$ equal to $0$ .

$x^{2} + 6 x + 36 = 0$

Solve $x^{2} + 6 x + 36 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 6$ , and $c = 36$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot 36)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 36$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 36}}{2 \cdot 1}$

Multiply $- 4$ by $36$ .

$x = \frac{- 6 \pm \sqrt{36 - 144}}{2 \cdot 1}$

Subtract $144$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 108}}{2 \cdot 1}$

Rewrite $- 108$ as $- 1 (108)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 108}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (108)}$ as $\sqrt{- 1} \cdot \sqrt{108}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{108}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{108}}{2 \cdot 1}$

Rewrite $108$ as $6^{2} \cdot 3$ .

Factor $36$ out of $108$ .

$x = \frac{- 6 \pm i \cdot \sqrt{36 (3)}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{6^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (6 \sqrt{3})}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{- 6 \pm 6 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 6 \pm 6 i \sqrt{3}}{2}$

Simplify $\frac{- 6 \pm 6 i \sqrt{3}}{2}$ .

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

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(x - 6) (x^{2} + 6 x + 36) = 0$ true.

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

Solve for x log base x of 216=3

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