Algebra Examples

3 \log (x) = \log (27)

$3 \log (x) = \log (27)$

Step 1

Simplify

3 \log (x)

$3 \log (x)$ by moving

3

$3$ inside the logarithm.

\log (x^{3}) = \log (27)

$\log (x^{3}) = \log (27)$

Step 2

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

x^{3} = 27

$x^{3} = 27$

Step 3

Solve for

x

$x$ .

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Step 3.1

Subtract

27

$27$ from both sides of the equation.

x^{3} - 27 = 0

$x^{3} - 27 = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Rewrite

27

$27$ as

3^{3}

$3^{3}$ .

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

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

Step 3.2.2

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})

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = x

$a = x$ and

b = 3

$b = 3$ .

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

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

Step 3.2.3

Simplify.

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Step 3.2.3.1

Move

3

$3$ to the left of

x

$x$ .

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

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

Step 3.2.3.2

Raise

3

$3$ to the power of

2

$2$ .

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

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

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

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

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

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

Step 3.3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

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

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

Step 3.4

Set

x - 3

$x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.4.1

Set

x - 3

$x - 3$ equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

Step 3.4.2

Add

3

$3$ to both sides of the equation.

x = 3

$x = 3$

x = 3

$x = 3$

Step 3.5

Set

x^{2} + 3 x + 9

$x^{2} + 3 x + 9$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.5.1

Set

x^{2} + 3 x + 9

$x^{2} + 3 x + 9$ equal to

0

$0$ .

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

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

Step 3.5.2

Solve

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

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

x

$x$ .

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Step 3.5.2.1

Use the quadratic formula to find the solutions.

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

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

Step 3.5.2.2

Substitute the values

a = 1

$a = 1$ ,

b = 3

$b = 3$ , and

c = 9

$c = 9$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 3.5.2.3

Simplify.

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Step 3.5.2.3.1

Simplify the numerator.

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Step 3.5.2.3.1.1

Raise

3

$3$ to the power of

2

$2$ .

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

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

Step 3.5.2.3.1.2

Multiply

- 4 \cdot 1 \cdot 9

$- 4 \cdot 1 \cdot 9$ .

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Step 3.5.2.3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 3.5.2.3.1.2.2

Multiply

- 4

$- 4$ by

9

$9$ .

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

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

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

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

Step 3.5.2.3.1.3

Subtract

36

$36$ from

9

$9$ .

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

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

Step 3.5.2.3.1.4

Rewrite

- 27

$- 27$ as

- 1 (27)

$- 1 (27)$ .

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

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

Step 3.5.2.3.1.5

Rewrite

\sqrt{- 1 (27)}

$\sqrt{- 1 (27)}$ as

\sqrt{- 1} \cdot \sqrt{27}

$\sqrt{- 1} \cdot \sqrt{27}$ .

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

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

Step 3.5.2.3.1.6

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

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

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

Step 3.5.2.3.1.7

Rewrite

27

$27$ as

3^{2} \cdot 3

$3^{2} \cdot 3$ .

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Step 3.5.2.3.1.7.1

Factor

9

$9$ out of

27

$27$ .

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

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

Step 3.5.2.3.1.7.2

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

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

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

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

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

Step 3.5.2.3.1.8

Pull terms out from under the radical.

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

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

Step 3.5.2.3.1.9

Move

3

$3$ to the left of

i

$i$ .

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

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

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

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

Step 3.5.2.3.2

Multiply

2

$2$ by

1

$1$ .

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

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

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

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

Step 3.5.2.4

The final answer is the combination of both solutions.

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

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

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

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

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

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

Step 3.6

The final solution is all the values that make

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

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

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

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

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

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