Precalculus Examples

ln (x - 4) + ln (x + 5) = 1

$\ln (x - 4) + \ln (x + 5) = 1$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms,

{log}_{b} (x) + {log}_{b} (y) = {log}_{b} (x y)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

ln ((x - 4) (x + 5)) = 1

$\ln ((x - 4) (x + 5)) = 1$

Step 1.2

Expand

(x - 4) (x + 5)

$(x - 4) (x + 5)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

ln (x (x + 5) - 4 (x + 5)) = 1

$\ln (x (x + 5) - 4 (x + 5)) = 1$

Step 1.2.2

Apply the distributive property.

ln (x \cdot x + x \cdot 5 - 4 (x + 5)) = 1

$\ln (x \cdot x + x \cdot 5 - 4 (x + 5)) = 1$

Step 1.2.3

Apply the distributive property.

ln (x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5) = 1

$\ln (x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5) = 1$

ln (x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5) = 1

$\ln (x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5) = 1$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply

x

$x$ by

x

$x$ .

ln (x^{2} + x \cdot 5 - 4 x - 4 \cdot 5) = 1

$\ln (x^{2} + x \cdot 5 - 4 x - 4 \cdot 5) = 1$

Step 1.3.1.2

Move

5

$5$ to the left of

x

$x$ .

ln (x^{2} + 5 \cdot x - 4 x - 4 \cdot 5) = 1

$\ln (x^{2} + 5 \cdot x - 4 x - 4 \cdot 5) = 1$

Step 1.3.1.3

Multiply

- 4

$- 4$ by

5

$5$ .

ln (x^{2} + 5 x - 4 x - 20) = 1

$\ln (x^{2} + 5 x - 4 x - 20) = 1$

ln (x^{2} + 5 x - 4 x - 20) = 1

$\ln (x^{2} + 5 x - 4 x - 20) = 1$

Step 1.3.2

Subtract

4 x

$4 x$ from

5 x

$5 x$ .

ln (x^{2} + x - 20) = 1

$\ln (x^{2} + x - 20) = 1$

ln (x^{2} + x - 20) = 1

$\ln (x^{2} + x - 20) = 1$

ln (x^{2} + x - 20) = 1

$\ln (x^{2} + x - 20) = 1$

Step 2

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

e^{ln (x^{2} + x - 20)} = e^{1}

$e^{\ln (x^{2} + x - 20)} = e^{1}$

Step 3

Rewrite

ln (x^{2} + x - 20) = 1

$\ln (x^{2} + x - 20) = 1$ 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

$b^{y} = x$ .

e^{1} = x^{2} + x - 20

$e^{1} = x^{2} + x - 20$

Step 4

Solve for

x

$x$ .

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Step 4.1

Rewrite the equation as

x^{2} + x - 20 = e^{1}

$x^{2} + x - 20 = e^{1}$ .

x^{2} + x - 20 = e

$x^{2} + x - 20 = e$

Step 4.2

Subtract

e

$e$ from both sides of the equation.

x^{2} + x - 20 - e = 0

$x^{2} + x - 20 - e = 0$

Step 4.3

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 4.4

Substitute the values

a = 1

$a = 1$ ,

b = 1

$b = 1$ , and

c = - 20 - e

$c = - 20 - e$ into the quadratic formula and solve for

x

$x$ .

\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot (- 20 - e))}}{2 \cdot 1}

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot (- 20 - e))}}{2 \cdot 1}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

One to any power is one.

x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 20 - e)}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 20 - e)}}{2 \cdot 1}$

Step 4.5.1.2

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (- 20 - e)}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (- 20 - e)}}{2 \cdot 1}$

Step 4.5.1.3

Apply the distributive property.

x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 20 - 4 (- e)}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 20 - 4 (- e)}}{2 \cdot 1}$

Step 4.5.1.4

Multiply

- 4

$- 4$ by

- 20

$- 20$ .

x = \frac{- 1 \pm \sqrt{1 + 80 - 4 (- e)}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{1 + 80 - 4 (- e)}}{2 \cdot 1}$

Step 4.5.1.5

Multiply

- 1

$- 1$ by

- 4

$- 4$ .

x = \frac{- 1 \pm \sqrt{1 + 80 + 4 e}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{1 + 80 + 4 e}}{2 \cdot 1}$

Step 4.5.1.6

Add

1

$1$ and

80

$80$ .

x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2 \cdot 1}$

x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2 \cdot 1}$

Step 4.5.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2}

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2}$

x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2}

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2}$

Step 4.6

The final answer is the combination of both solutions.

x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}$

x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}$

Step 5

Exclude the solutions that do not make

ln (x - 4) + ln (x + 5) = 1

$\ln (x - 4) + \ln (x + 5) = 1$ true.

x = - \frac{1 - \sqrt{81 + 4 e}}{2}

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

x = - \frac{1 - \sqrt{81 + 4 e}}{2}

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}$

Decimal Form:

x = 4.29252353 \dots

$x = 4.29252353 \dots$