Precalculus Examples

$\ln (\frac{(2 x + 1) (x - 9)}{x^{2}}) = 0$

Step 1

Rewrite $\ln (\frac{(2 x + 1) (x - 9)}{x^{2}}) = 0$ 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 $b^{y} = x$ .

$e^{0} = \frac{(2 x + 1) (x - 9)}{x^{2}}$

Step 2

Cross multiply to remove the fraction.

$(2 x + 1) (x - 9) = e^{0} (x^{2})$

Step 3

Simplify $e^{0} (x^{2})$ .

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Step 3.1

Anything raised to $0$ is $1$ .

$(2 x + 1) (x - 9) = 1 x^{2}$

Step 3.2

Multiply $x^{2}$ by $1$ .

$(2 x + 1) (x - 9) = x^{2}$

Step 4

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

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Step 4.1

Subtract $x^{2}$ from both sides of the equation.

$(2 x + 1) (x - 9) - x^{2} = 0$

Step 4.2

Simplify each term.

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Step 4.2.1

Expand $(2 x + 1) (x - 9)$ using the FOIL Method.

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Step 4.2.1.1

Apply the distributive property.

$2 x (x - 9) + 1 (x - 9) - x^{2} = 0$

Step 4.2.1.2

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 9 + 1 (x - 9) - x^{2} = 0$

Step 4.2.1.3

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 9 + 1 x + 1 \cdot - 9 - x^{2} = 0$

Step 4.2.2

Simplify and combine like terms.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 4.2.2.1.1.1

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 9 + 1 x + 1 \cdot - 9 - x^{2} = 0$

Step 4.2.2.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 9 + 1 x + 1 \cdot - 9 - x^{2} = 0$

Step 4.2.2.1.2

Multiply $- 9$ by $2$ .

$2 x^{2} - 18 x + 1 x + 1 \cdot - 9 - x^{2} = 0$

Step 4.2.2.1.3

Multiply $x$ by $1$ .

$2 x^{2} - 18 x + x + 1 \cdot - 9 - x^{2} = 0$

Step 4.2.2.1.4

Multiply $- 9$ by $1$ .

$2 x^{2} - 18 x + x - 9 - x^{2} = 0$

Step 4.2.2.2

Add $- 18 x$ and $x$ .

$2 x^{2} - 17 x - 9 - x^{2} = 0$

Step 4.3

Subtract $x^{2}$ from $2 x^{2}$ .

$x^{2} - 17 x - 9 = 0$

Step 5

Add $9$ to both sides of the equation.

$x^{2} - 17 x = 9$

Step 6

Subtract $9$ from both sides of the equation.

$x^{2} - 17 x - 9 = 0$

Step 7

Use the quadratic formula to find the solutions.

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

Step 8

Substitute the values $a = 1$ , $b = - 17$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (1 \cdot - 9)}}{2 \cdot 1}$

Step 9

Simplify.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 1 \cdot - 9}}{2 \cdot 1}$

Step 9.1.2

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

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Step 9.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot - 9}}{2 \cdot 1}$

Step 9.1.2.2

Multiply $- 4$ by $- 9$ .

$x = \frac{17 \pm \sqrt{289 + 36}}{2 \cdot 1}$

Step 9.1.3

Add $289$ and $36$ .

$x = \frac{17 \pm \sqrt{325}}{2 \cdot 1}$

Step 9.1.4

Rewrite $325$ as $5^{2} \cdot 13$ .

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Step 9.1.4.1

Factor $25$ out of $325$ .

$x = \frac{17 \pm \sqrt{25 (13)}}{2 \cdot 1}$

Step 9.1.4.2

Rewrite $25$ as $5^{2}$ .

$x = \frac{17 \pm \sqrt{5^{2} \cdot 13}}{2 \cdot 1}$

Step 9.1.5

Pull terms out from under the radical.

$x = \frac{17 \pm 5 \sqrt{13}}{2 \cdot 1}$

Step 9.2

Multiply $2$ by $1$ .

$x = \frac{17 \pm 5 \sqrt{13}}{2}$

Step 10

The final answer is the combination of both solutions.

$x = \frac{17 + 5 \sqrt{13}}{2}, \frac{17 - 5 \sqrt{13}}{2}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$x = \frac{17 + 5 \sqrt{13}}{2}, \frac{17 - 5 \sqrt{13}}{2}$

Decimal Form:

$x = 17.51387818 \dots, - 0.51387818 \dots$