Precalculus Examples

$\log (x + 8) + \log (x - 1) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((x + 8) (x - 1)) = 1$

Step 1.2

Expand $(x + 8) (x - 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log (x (x - 1) + 8 (x - 1)) = 1$

Step 1.2.2

Apply the distributive property.

$\log (x \cdot x + x \cdot - 1 + 8 (x - 1)) = 1$

Step 1.2.3

Apply the distributive property.

$\log (x \cdot x + x \cdot - 1 + 8 x + 8 \cdot - 1) = 1$

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 $x$ .

$\log (x^{2} + x \cdot - 1 + 8 x + 8 \cdot - 1) = 1$

Step 1.3.1.2

Move $- 1$ to the left of $x$ .

$\log (x^{2} - 1 \cdot x + 8 x + 8 \cdot - 1) = 1$

Step 1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\log (x^{2} - x + 8 x + 8 \cdot - 1) = 1$

Step 1.3.1.4

Multiply $8$ by $- 1$ .

$\log (x^{2} - x + 8 x - 8) = 1$

Step 1.3.2

Add $- x$ and $8 x$ .

$\log (x^{2} + 7 x - 8) = 1$

Step 2

Rewrite $\log (x^{2} + 7 x - 8) = 1$ 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 $b^{y} = x$ .

$10^{1} = x^{2} + 7 x - 8$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} + 7 x - 8 = 10^{1}$ .

$x^{2} + 7 x - 8 = 10$

Step 3.2

Subtract $10$ from both sides of the equation.

$x^{2} + 7 x - 8 - 10 = 0$

Step 3.3

Subtract $10$ from $- 8$ .

$x^{2} + 7 x - 18 = 0$

Step 3.4

Factor $x^{2} + 7 x - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Step 3.4.2

Write the factored form using these integers.

$(x - 2) (x + 9) = 0$

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 $0$ .

$x - 2 = 0$

$x + 9 = 0$

Step 3.6

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

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Step 3.6.1

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

$x - 2 = 0$

Step 3.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 3.7

Set $x + 9$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Step 3.7.2

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 3.8

The final solution is all the values that make $(x - 2) (x + 9) = 0$ true.

$x = 2, - 9$

Step 4

Exclude the solutions that do not make $\log (x + 8) + \log (x - 1) = 1$ true.

$x = 2$