Algebra Examples

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

Step 1

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 $b^{y} = x$ .

$x^{3} = 27$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $27$ from both sides of the equation.

$x^{3} - 27 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Rewrite $27$ as $3^{3}$ .

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

Step 2.2.2

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 $b = 3$ .

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.3

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

$x - 3 = 0$

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

Step 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

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

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Step 2.5.2.1

Use the quadratic formula to find the solutions.

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

Step 2.5.2.2

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

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

Step 2.5.2.3

Simplify.

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Step 2.5.2.3.1

Simplify the numerator.

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Step 2.5.2.3.1.1

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

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

Step 2.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 2.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.2.3.1.2.2

Multiply $- 4$ by $9$ .

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

Step 2.5.2.3.1.3

Subtract $36$ from $9$ .

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

Step 2.5.2.3.1.4

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

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

Step 2.5.2.3.1.5

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

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

Step 2.5.2.3.1.6

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

$x = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Step 2.5.2.3.1.7

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

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Step 2.5.2.3.1.7.1

Factor $9$ out of $27$ .

$x = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Step 2.5.2.3.1.7.2

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

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

Step 2.5.2.3.1.8

Pull terms out from under the radical.

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

Step 2.5.2.3.1.9

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

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

Step 2.5.2.3.2

Multiply $2$ by $1$ .

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

Step 2.5.2.4

The final answer is the combination of both solutions.

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

Step 2.6

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

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