Precalculus Examples

$\log_{4} (x^{2} + 3 x) - \log_{4} (x + 5) = 1$

Step 1

Simplify the left side.

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Step 1.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{4} (\frac{x^{2} + 3 x}{x + 5}) = 1$

Step 1.2

Factor $x$ out of $x^{2} + 3 x$ .

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Step 1.2.1

Factor $x$ out of $x^{2}$ .

$\log_{4} (\frac{x \cdot x + 3 x}{x + 5}) = 1$

Step 1.2.2

Factor $x$ out of $3 x$ .

$\log_{4} (\frac{x \cdot x + x \cdot 3}{x + 5}) = 1$

Step 1.2.3

Factor $x$ out of $x \cdot x + x \cdot 3$ .

$\log_{4} (\frac{x (x + 3)}{x + 5}) = 1$

Step 2

Rewrite $\log_{4} (\frac{x (x + 3)}{x + 5}) = 1$ 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 $b^{y} = x$ .

$4^{1} = \frac{x (x + 3)}{x + 5}$

Step 3

Cross multiply to remove the fraction.

$x (x + 3) = 4 (x + 5)$

Step 4

Simplify $4 (x + 5)$ .

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Step 4.1

Apply the distributive property.

$x (x + 3) = 4 x + 4 \cdot 5$

Step 4.2

Multiply $4$ by $5$ .

$x (x + 3) = 4 x + 20$

Step 5

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

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Step 5.1

Subtract $4 x$ from both sides of the equation.

$x (x + 3) - 4 x = 20$

Step 5.2

Simplify each term.

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Step 5.2.1

Apply the distributive property.

$x \cdot x + x \cdot 3 - 4 x = 20$

Step 5.2.2

Multiply $x$ by $x$ .

$x^{2} + x \cdot 3 - 4 x = 20$

Step 5.2.3

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

$x^{2} + 3 x - 4 x = 20$

Step 5.3

Subtract $4 x$ from $3 x$ .

$x^{2} - x = 20$

Step 6

Factor $x$ out of $x^{2} - x$ .

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Step 6.1

Factor $x$ out of $x^{2}$ .

$x \cdot x - x = 20$

Step 6.2

Factor $x$ out of $- x$ .

$x \cdot x + x \cdot - 1 = 20$

Step 6.3

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

$x (x - 1) = 20$

Step 7

Simplify $x (x - 1)$ .

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Step 7.1

Simplify by multiplying through.

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Step 7.1.1

Apply the distributive property.

$x \cdot x + x \cdot - 1 = 20$

Step 7.1.2

Simplify the expression.

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Step 7.1.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 = 20$

Step 7.1.2.2

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

$x^{2} - 1 \cdot x = 20$

Step 7.2

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

$x^{2} - x = 20$

Step 8

Subtract $20$ from both sides of the equation.

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

Step 9

Factor $x^{2} - x - 20$ using the AC method.

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Step 9.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 9.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 10

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

$x - 5 = 0$

$x + 4 = 0$

Step 11

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 11.2

Add $5$ to both sides of the equation.

$x = 5$

Step 12

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 12.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 13

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$