Precalculus Examples

\log_{x} (125) = 3

$\log_{x} (125) = 3$

Step 1

Rewrite

\log_{x} (125) = 3

$\log_{x} (125) = 3$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

x^{3} = 125

$x^{3} = 125$

Step 2

Solve for

x

$x$ .

Tap for more steps...

Step 2.1

Subtract

125

$125$ from both sides of the equation.

x^{3} - 125 = 0

$x^{3} - 125 = 0$

Step 2.2

Factor the left side of the equation.

Tap for more steps...

Step 2.2.1

Rewrite

125

$125$ as

5^{3}

$5^{3}$ .

x^{3} - 5^{3} = 0

$x^{3} - 5^{3} = 0$

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})

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

a = x

$a = x$ and

b = 5

$b = 5$ .

(x - 5) (x^{2} + x \cdot 5 + 5^{2}) = 0

$(x - 5) (x^{2} + x \cdot 5 + 5^{2}) = 0$

Step 2.2.3

Simplify.

Tap for more steps...

Step 2.2.3.1

Move

5

$5$ to the left of

x

$x$ .

(x - 5) (x^{2} + 5 x + 5^{2}) = 0

$(x - 5) (x^{2} + 5 x + 5^{2}) = 0$

Step 2.2.3.2

Raise

5

$5$ to the power of

2

$2$ .

(x - 5) (x^{2} + 5 x + 25) = 0

$(x - 5) (x^{2} + 5 x + 25) = 0$

(x - 5) (x^{2} + 5 x + 25) = 0

$(x - 5) (x^{2} + 5 x + 25) = 0$

(x - 5) (x^{2} + 5 x + 25) = 0

$(x - 5) (x^{2} + 5 x + 25) = 0$

Step 2.3

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 - 5 = 0

$x - 5 = 0$

x^{2} + 5 x + 25 = 0

$x^{2} + 5 x + 25 = 0$

Step 2.4

Set

x - 5

$x - 5$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 2.4.1

Set

x - 5

$x - 5$ equal to

0

$0$ .

x - 5 = 0

$x - 5 = 0$

Step 2.4.2

Add

5

$5$ to both sides of the equation.

x = 5

$x = 5$

x = 5

$x = 5$

Step 2.5

Set

x^{2} + 5 x + 25

$x^{2} + 5 x + 25$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 2.5.1

Set

x^{2} + 5 x + 25

$x^{2} + 5 x + 25$ equal to

0

$0$ .

x^{2} + 5 x + 25 = 0

$x^{2} + 5 x + 25 = 0$

Step 2.5.2

Solve

x^{2} + 5 x + 25 = 0

$x^{2} + 5 x + 25 = 0$ for

x

$x$ .

Tap for more steps...

Step 2.5.2.1

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 2.5.2.2

Substitute the values

a = 1

$a = 1$ ,

b = 5

$b = 5$ , and

c = 25

$c = 25$ into the quadratic formula and solve for

x

$x$ .

\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Step 2.5.2.3

Simplify.

Tap for more steps...

Step 2.5.2.3.1

Simplify the numerator.

Tap for more steps...

Step 2.5.2.3.1.1

Raise

5

$5$ to the power of

2

$2$ .

x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Step 2.5.2.3.1.2

Multiply

- 4 \cdot 1 \cdot 25

$- 4 \cdot 1 \cdot 25$ .

Tap for more steps...

Step 2.5.2.3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 25}}{2 \cdot 1}$

Step 2.5.2.3.1.2.2

Multiply

- 4

$- 4$ by

25

$25$ .

x = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 1}$

x = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{25 - 100}}{2 \cdot 1}$

Step 2.5.2.3.1.3

Subtract

100

$100$ from

25

$25$ .

x = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{- 75}}{2 \cdot 1}$

Step 2.5.2.3.1.4

Rewrite

- 75

$- 75$ as

- 1 (75)

$- 1 (75)$ .

x = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 1}$

Step 2.5.2.3.1.5

Rewrite

\sqrt{- 1 (75)}

$\sqrt{- 1 (75)}$ as

\sqrt{- 1} \cdot \sqrt{75}

$\sqrt{- 1} \cdot \sqrt{75}$ .

x = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 1}

$x = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 1}$

Step 2.5.2.3.1.6

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

x = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 1}

$x = \frac{- 5 \pm i \cdot \sqrt{75}}{2 \cdot 1}$

Step 2.5.2.3.1.7

Rewrite

75

$75$ as

5^{2} \cdot 3

$5^{2} \cdot 3$ .

Tap for more steps...

Step 2.5.2.3.1.7.1

Factor

25

$25$ out of

75

$75$ .

x = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 1}

$x = \frac{- 5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 1}$

Step 2.5.2.3.1.7.2

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

x = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}

$x = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}$

x = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}

$x = \frac{- 5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}$

Step 2.5.2.3.1.8

Pull terms out from under the radical.

x = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 1}

$x = \frac{- 5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 1}$

Step 2.5.2.3.1.9

Move

5

$5$ to the left of

i

$i$ .

x = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 1}$

x = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2 \cdot 1}$

Step 2.5.2.3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 5 \pm 5 i \sqrt{3}}{2}

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2}$

x = \frac{- 5 \pm 5 i \sqrt{3}}{2}

$x = \frac{- 5 \pm 5 i \sqrt{3}}{2}$

Step 2.5.2.4

The final answer is the combination of both solutions.

x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}

$x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}

$x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}

$x = - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

Step 2.6

The final solution is all the values that make

(x - 5) (x^{2} + 5 x + 25) = 0

$(x - 5) (x^{2} + 5 x + 25) = 0$ true.

x = 5, - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}

$x = 5, - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$

x = 5, - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}

$x = 5, - \frac{5 - 5 i \sqrt{3}}{2}, - \frac{5 + 5 i \sqrt{3}}{2}$