Precalculus Examples

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

Step 1

Simplify $2 \ln (x + 2)$ by moving $2$ inside the logarithm.

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

Step 2

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

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

Step 3

Solve for $x$ .

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Step 3.1

Move all terms containing $x$ to the left side of the equation.

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Step 3.1.1

Subtract $10 x$ from both sides of the equation.

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

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

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Step 3.1.2.2.1

Apply the distributive property.

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

Step 3.1.2.2.2

Apply the distributive property.

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

Step 3.1.2.2.3

Apply the distributive property.

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

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 $x$ by $x$ .

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

Step 3.1.2.3.1.2

Move $2$ to the left of $x$ .

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

Step 3.1.2.3.1.3

Multiply $2$ by $2$ .

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

Step 3.1.2.3.2

Add $2 x$ and $2 x$ .

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

Step 3.1.3

Subtract $10 x$ from $4 x$ .

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

Step 3.2

Use the quadratic formula to find the solutions.

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

Step 3.3

Substitute the values $a = 1$ , $b = - 6$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

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

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

Step 3.4.1.2

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

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Step 3.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.4.1.2.2

Multiply $- 4$ by $4$ .

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

Step 3.4.1.3

Subtract $16$ from $36$ .

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

Step 3.4.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 3.4.1.4.1

Factor $4$ out of $20$ .

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

Step 3.4.1.4.2

Rewrite $4$ as $2^{2}$ .

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

Step 3.4.1.5

Pull terms out from under the radical.

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

Step 3.4.2

Multiply $2$ by $1$ .

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

Step 3.4.3

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

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

Step 3.5

The final answer is the combination of both solutions.

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 5.23606797 \dots, 0.76393202 \dots$