Precalculus Examples

$\log (2 x + 1) = \log (x - 3) + \log (x + 5)$

Step 1

Simplify the right side.

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Step 1.1

Simplify $\log (x - 3) + \log (x + 5)$ .

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Step 1.1.1

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

$\log (2 x + 1) = \log ((x - 3) (x + 5))$

Step 1.1.2

Expand $(x - 3) (x + 5)$ using the FOIL Method.

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Step 1.1.2.1

Apply the distributive property.

$\log (2 x + 1) = \log (x (x + 5) - 3 (x + 5))$

Step 1.1.2.2

Apply the distributive property.

$\log (2 x + 1) = \log (x \cdot x + x \cdot 5 - 3 (x + 5))$

Step 1.1.2.3

Apply the distributive property.

$\log (2 x + 1) = \log (x \cdot x + x \cdot 5 - 3 x - 3 \cdot 5)$

Step 1.1.3

Simplify and combine like terms.

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Step 1.1.3.1

Simplify each term.

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Step 1.1.3.1.1

Multiply $x$ by $x$ .

$\log (2 x + 1) = \log (x^{2} + x \cdot 5 - 3 x - 3 \cdot 5)$

Step 1.1.3.1.2

Move $5$ to the left of $x$ .

$\log (2 x + 1) = \log (x^{2} + 5 \cdot x - 3 x - 3 \cdot 5)$

Step 1.1.3.1.3

Multiply $- 3$ by $5$ .

$\log (2 x + 1) = \log (x^{2} + 5 x - 3 x - 15)$

Step 1.1.3.2

Subtract $3 x$ from $5 x$ .

$\log (2 x + 1) = \log (x^{2} + 2 x - 15)$

Step 2

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

$2 x + 1 = x^{2} + 2 x - 15$

Step 3

Solve for $x$ .

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Step 3.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 2 x - 15 = 2 x + 1$

Step 3.2

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

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Step 3.2.1

Subtract $2 x$ from both sides of the equation.

$x^{2} + 2 x - 15 - 2 x = 1$

Step 3.2.2

Combine the opposite terms in $x^{2} + 2 x - 15 - 2 x$ .

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Step 3.2.2.1

Subtract $2 x$ from $2 x$ .

$x^{2} + 0 - 15 = 1$

Step 3.2.2.2

Add $x^{2}$ and $0$ .

$x^{2} - 15 = 1$

Step 3.3

Move all terms not containing $x$ to the right side of the equation.

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Step 3.3.1

Add $15$ to both sides of the equation.

$x^{2} = 1 + 15$

Step 3.3.2

Add $1$ and $15$ .

$x^{2} = 16$

Step 3.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Step 3.5

Simplify $\pm \sqrt{16}$ .

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Step 3.5.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Step 3.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

Step 3.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Step 3.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

Step 3.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$

Step 4

Exclude the solutions that do not make $\log (2 x + 1) = \log (x - 3) + \log (x + 5)$ true.

$x = 4$