Precalculus Examples

$\ln (x - 2) + \ln (x + 3) = 1$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln ((x - 2) (x + 3)) = 1$

Step 1.2

Expand $(x - 2) (x + 3)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\ln (x (x + 3) - 2 (x + 3)) = 1$

Step 1.2.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot 3 - 2 (x + 3)) = 1$

Step 1.2.3

Apply the distributive property.

$\ln (x \cdot x + x \cdot 3 - 2 x - 2 \cdot 3) = 1$

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 $x$ by $x$ .

$\ln (x^{2} + x \cdot 3 - 2 x - 2 \cdot 3) = 1$

Step 1.3.1.2

Move $3$ to the left of $x$ .

$\ln (x^{2} + 3 \cdot x - 2 x - 2 \cdot 3) = 1$

Step 1.3.1.3

Multiply $- 2$ by $3$ .

$\ln (x^{2} + 3 x - 2 x - 6) = 1$

Step 1.3.2

Subtract $2 x$ from $3 x$ .

$\ln (x^{2} + x - 6) = 1$

Step 2

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x^{2} + x - 6)} = e^{1}$

Step 3

Rewrite $\ln (x^{2} + x - 6) = 1$ 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 $b^{y} = x$ .

$e^{1} = x^{2} + x - 6$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $x^{2} + x - 6 = e^{1}$ .

$x^{2} + x - 6 = e$

Step 4.2

Subtract $e$ from both sides of the equation.

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

Step 4.3

Use the quadratic formula to find the solutions.

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

Step 4.4

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

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

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.

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

Step 4.5.1.2

Multiply $- 4$ by $1$ .

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

Step 4.5.1.3

Apply the distributive property.

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

Step 4.5.1.4

Multiply $- 4$ by $- 6$ .

$x = \frac{- 1 \pm \sqrt{1 + 24 - 4 (- e)}}{2 \cdot 1}$

Step 4.5.1.5

Multiply $- 1$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 24 + 4 e}}{2 \cdot 1}$

Step 4.5.1.6

Add $1$ and $24$ .

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

Step 4.5.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{25 + 4 e}}{2}$

Step 4.6

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{25 + 4 e}}{2}, - \frac{1 + \sqrt{25 + 4 e}}{2}$

Step 5

Exclude the solutions that do not make $\ln (x - 2) + \ln (x + 3) = 1$ true.

$x = - \frac{1 - \sqrt{25 + 4 e}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{25 + 4 e}}{2}$

Decimal Form:

$x = 2.49470897 \dots$