Precalculus Examples

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

Step 1

Simplify the left side.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

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

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

Step 3

Cross multiply to remove the fraction.

$(x - 7) (x - 4) = 5 (x)$

Step 4

Multiply $5$ by $x$ .

$(x - 7) (x - 4) = 5 x$

Step 5

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

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Step 5.1

Subtract $5 x$ from both sides of the equation.

$(x - 7) (x - 4) - 5 x = 0$

Step 5.2

Simplify each term.

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Step 5.2.1

Expand $(x - 7) (x - 4)$ using the FOIL Method.

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Step 5.2.1.1

Apply the distributive property.

$x (x - 4) - 7 (x - 4) - 5 x = 0$

Step 5.2.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 4 - 7 (x - 4) - 5 x = 0$

Step 5.2.1.3

Apply the distributive property.

$x \cdot x + x \cdot - 4 - 7 x - 7 \cdot - 4 - 5 x = 0$

Step 5.2.2

Simplify and combine like terms.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply $x$ by $x$ .

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

Step 5.2.2.1.2

Move $- 4$ to the left of $x$ .

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

Step 5.2.2.1.3

Multiply $- 7$ by $- 4$ .

$x^{2} - 4 x - 7 x + 28 - 5 x = 0$

Step 5.2.2.2

Subtract $7 x$ from $- 4 x$ .

$x^{2} - 11 x + 28 - 5 x = 0$

Step 5.3

Subtract $5 x$ from $- 11 x$ .

$x^{2} - 16 x + 28 = 0$

Step 6

Factor $x^{2} - 16 x + 28$ using the AC method.

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Step 6.1

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 $28$ and whose sum is $- 16$ .

$- 14, - 2$

Step 6.2

Write the factored form using these integers.

$(x - 14) (x - 2) = 0$

Step 7

Simplify $(x - 14) (x - 2)$ .

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Step 7.1

Expand $(x - 14) (x - 2)$ using the FOIL Method.

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Step 7.1.1

Apply the distributive property.

$x (x - 2) - 14 (x - 2) = 0$

Step 7.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 2 - 14 (x - 2) = 0$

Step 7.1.3

Apply the distributive property.

$x \cdot x + x \cdot - 2 - 14 x - 14 \cdot - 2 = 0$

Step 7.2

Simplify and combine like terms.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2 - 14 x - 14 \cdot - 2 = 0$

Step 7.2.1.2

Move $- 2$ to the left of $x$ .

$x^{2} - 2 \cdot x - 14 x - 14 \cdot - 2 = 0$

Step 7.2.1.3

Multiply $- 14$ by $- 2$ .

$x^{2} - 2 x - 14 x + 28 = 0$

Step 7.2.2

Subtract $14 x$ from $- 2 x$ .

$x^{2} - 16 x + 28 = 0$

Step 8

Factor $x^{2} - 16 x + 28$ using the AC method.

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Step 8.1

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 $28$ and whose sum is $- 16$ .

$- 14, - 2$

Step 8.2

Write the factored form using these integers.

$(x - 14) (x - 2) = 0$

Step 9

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

$x - 14 = 0$

$x - 2 = 0$

Step 10

Set $x - 14$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $x - 14$ equal to $0$ .

$x - 14 = 0$

Step 10.2

Add $14$ to both sides of the equation.

$x = 14$

Step 11

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

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Step 11.1

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

$x - 2 = 0$

Step 11.2

Add $2$ to both sides of the equation.

$x = 2$

Step 12

The final solution is all the values that make $(x - 14) (x - 2) = 0$ true.

$x = 14, 2$

Step 13

Exclude the solutions that do not make $\log_{5} (x - 7) + \log_{5} (x - 4) - \log_{5} (x) = 1$ true.

$x = 14$