Precalculus Examples

2 \ln (x + 2) = \ln (10 x)

$2 \ln (x + 2) = \ln (10 x)$

Step 1

Simplify

2 \ln (x + 2)

$2 \ln (x + 2)$ by moving

2

$2$ inside the logarithm.

\ln ({(x + 2)}^{2}) = \ln (10 x)

$\ln ({(x + 2)}^{2}) = \ln (10 x)$

Step 2

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

{(x + 2)}^{2} = 10 x

${(x + 2)}^{2} = 10 x$

Step 3

Solve for

x

$x$ .

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Step 3.1

Move all terms containing

x

$x$ to the left side of the equation.

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Step 3.1.1

Subtract

10 x

$10 x$ from both sides of the equation.

{(x + 2)}^{2} - 10 x = 0

${(x + 2)}^{2} - 10 x = 0$

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

Rewrite

{(x + 2)}^{2}

${(x + 2)}^{2}$ as

(x + 2) (x + 2)

$(x + 2) (x + 2)$ .

(x + 2) (x + 2) - 10 x = 0

$(x + 2) (x + 2) - 10 x = 0$

Step 3.1.2.2

Expand

(x + 2) (x + 2)

$(x + 2) (x + 2)$ using the FOIL Method.

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Step 3.1.2.2.1

Apply the distributive property.

x (x + 2) + 2 (x + 2) - 10 x = 0

$x (x + 2) + 2 (x + 2) - 10 x = 0$

Step 3.1.2.2.2

Apply the distributive property.

x \cdot x + x \cdot 2 + 2 (x + 2) - 10 x = 0

$x \cdot x + x \cdot 2 + 2 (x + 2) - 10 x = 0$

Step 3.1.2.2.3

Apply the distributive property.

x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0

$x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0$

x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0

$x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0$

Step 3.1.2.3

Simplify and combine like terms.

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Step 3.1.2.3.1

Simplify each term.

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Step 3.1.2.3.1.1

Multiply

x

$x$ by

x

$x$ .

x^{2} + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0

$x^{2} + x \cdot 2 + 2 x + 2 \cdot 2 - 10 x = 0$

Step 3.1.2.3.1.2

Move

2

$2$ to the left of

x

$x$ .

x^{2} + 2 \cdot x + 2 x + 2 \cdot 2 - 10 x = 0

$x^{2} + 2 \cdot x + 2 x + 2 \cdot 2 - 10 x = 0$

Step 3.1.2.3.1.3

Multiply

2

$2$ by

2

$2$ .

x^{2} + 2 x + 2 x + 4 - 10 x = 0

$x^{2} + 2 x + 2 x + 4 - 10 x = 0$

x^{2} + 2 x + 2 x + 4 - 10 x = 0

$x^{2} + 2 x + 2 x + 4 - 10 x = 0$

Step 3.1.2.3.2

Add

2 x

$2 x$ and

2 x

$2 x$ .

x^{2} + 4 x + 4 - 10 x = 0

$x^{2} + 4 x + 4 - 10 x = 0$

x^{2} + 4 x + 4 - 10 x = 0

$x^{2} + 4 x + 4 - 10 x = 0$

x^{2} + 4 x + 4 - 10 x = 0

$x^{2} + 4 x + 4 - 10 x = 0$

Step 3.1.3

Subtract

10 x

$10 x$ from

4 x

$4 x$ .

x^{2} - 6 x + 4 = 0

$x^{2} - 6 x + 4 = 0$

x^{2} - 6 x + 4 = 0

$x^{2} - 6 x + 4 = 0$

Step 3.2

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 3.3

Substitute the values

a = 1

$a = 1$ ,

b = - 6

$b = - 6$ , and

c = 4

$c = 4$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise

- 6

$- 6$ to the power of

2

$2$ .

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

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

Step 3.4.1.2

Multiply

- 4 \cdot 1 \cdot 4

$- 4 \cdot 1 \cdot 4$ .

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Step 3.4.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 3.4.1.2.2

Multiply

- 4

$- 4$ by

4

$4$ .

x = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1}$

x = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1}$

Step 3.4.1.3

Subtract

16

$16$ from

36

$36$ .

x = \frac{6 \pm \sqrt{20}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{20}}{2 \cdot 1}$

Step 3.4.1.4

Rewrite

20

$20$ as

2^{2} \cdot 5

$2^{2} \cdot 5$ .

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Step 3.4.1.4.1

Factor

4

$4$ out of

20

$20$ .

x = \frac{6 \pm \sqrt{4 (5)}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Step 3.4.1.4.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

x = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

x = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}

$x = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Step 3.4.1.5

Pull terms out from under the radical.

x = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 1}

$x = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 1}$

x = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 1}

$x = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 1}$

Step 3.4.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{6 \pm 2 \sqrt{5}}{2}

$x = \frac{6 \pm 2 \sqrt{5}}{2}$

Step 3.4.3

Simplify

\frac{6 \pm 2 \sqrt{5}}{2}

$\frac{6 \pm 2 \sqrt{5}}{2}$ .

x = 3 \pm \sqrt{5}

$x = 3 \pm \sqrt{5}$

x = 3 \pm \sqrt{5}

$x = 3 \pm \sqrt{5}$

Step 3.5

The final answer is the combination of both solutions.

x = 3 + \sqrt{5}, 3 - \sqrt{5}

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

x = 3 + \sqrt{5}, 3 - \sqrt{5}

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

Step 4

The result can be shown in multiple forms.

Exact Form:

x = 3 + \sqrt{5}, 3 - \sqrt{5}

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

Decimal Form:

x = 5.23606797 \dots, 0.76393202 \dots

$x = 5.23606797 \dots, 0.76393202 \dots$