Trigonometry Examples

$\log_{7} (x - 5) + \log_{7} (x + 1) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{7} ((x - 5) (x + 1)) = 1$

Step 1.2

Expand $(x - 5) (x + 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{7} (x (x + 1) - 5 (x + 1)) = 1$

Step 1.2.2

Apply the distributive property.

$\log_{7} (x \cdot x + x \cdot 1 - 5 (x + 1)) = 1$

Step 1.2.3

Apply the distributive property.

$\log_{7} (x \cdot x + x \cdot 1 - 5 x - 5 \cdot 1) = 1$

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 $x$ .

$\log_{7} (x^{2} + x \cdot 1 - 5 x - 5 \cdot 1) = 1$

Step 1.3.1.2

Multiply $x$ by $1$ .

$\log_{7} (x^{2} + x - 5 x - 5 \cdot 1) = 1$

Step 1.3.1.3

Multiply $- 5$ by $1$ .

$\log_{7} (x^{2} + x - 5 x - 5) = 1$

Step 1.3.2

Subtract $5 x$ from $x$ .

$\log_{7} (x^{2} - 4 x - 5) = 1$

Step 2

Rewrite $\log_{7} (x^{2} - 4 x - 5) = 1$ 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 $b^{y} = x$ .

$7^{1} = x^{2} - 4 x - 5$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - 4 x - 5 = 7^{1}$ .

$x^{2} - 4 x - 5 = 7$

Step 3.2

Subtract $7$ from both sides of the equation.

$x^{2} - 4 x - 5 - 7 = 0$

Step 3.3

Subtract $7$ from $- 5$ .

$x^{2} - 4 x - 12 = 0$

Step 3.4

Factor $x^{2} - 4 x - 12$ using the AC method.

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Step 3.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 3.4.2

Write the factored form using these integers.

$(x - 6) (x + 2) = 0$

Step 3.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x + 2 = 0$

Step 3.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 3.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.7

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 3.7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.8

The final solution is all the values that make $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Step 4

Exclude the solutions that do not make $\log_{7} (x - 5) + \log_{7} (x + 1) = 1$ true.

$x = 6$