Precalculus Examples

$\log (x - 6) - \log (8 x - 40) = \log (\frac{1}{x})$

Step 1

Simplify the left side.

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Step 1.1

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

$\log (\frac{x - 6}{8 x - 40}) = \log (\frac{1}{x})$

Step 1.2

Factor $8$ out of $8 x - 40$ .

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Step 1.2.1

Factor $8$ out of $8 x$ .

$\log (\frac{x - 6}{8 (x) - 40}) = \log (\frac{1}{x})$

Step 1.2.2

Factor $8$ out of $- 40$ .

$\log (\frac{x - 6}{8 x + 8 \cdot - 5}) = \log (\frac{1}{x})$

Step 1.2.3

Factor $8$ out of $8 x + 8 \cdot - 5$ .

$\log (\frac{x - 6}{8 (x - 5)}) = \log (\frac{1}{x})$

Step 2

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

$\frac{x - 6}{8 (x - 5)} = \frac{1}{x}$

Step 3

Solve for $x$ .

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Step 3.1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(x - 6) x = 8 (x - 5) \cdot 1$

Step 3.2

Solve the equation for $x$ .

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Step 3.2.1

Simplify $(x - 6) x$ .

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Step 3.2.1.1

Rewrite.

$0 + 0 + (x - 6) x = 8 (x - 5) \cdot 1$

Step 3.2.1.2

Simplify by adding zeros.

$(x - 6) x = 8 (x - 5) \cdot 1$

Step 3.2.1.3

Apply the distributive property.

$x \cdot x - 6 x = 8 (x - 5) \cdot 1$

Step 3.2.1.4

Multiply $x$ by $x$ .

$x^{2} - 6 x = 8 (x - 5) \cdot 1$

Step 3.2.2

Simplify $8 (x - 5) \cdot 1$ .

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Step 3.2.2.1

Apply the distributive property.

$x^{2} - 6 x = (8 x + 8 \cdot - 5) \cdot 1$

Step 3.2.2.2

Multiply.

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Step 3.2.2.2.1

Multiply $8$ by $- 5$ .

$x^{2} - 6 x = (8 x - 40) \cdot 1$

Step 3.2.2.2.2

Multiply $8 x - 40$ by $1$ .

$x^{2} - 6 x = 8 x - 40$

Step 3.2.3

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

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Step 3.2.3.1

Subtract $8 x$ from both sides of the equation.

$x^{2} - 6 x - 8 x = - 40$

Step 3.2.3.2

Subtract $8 x$ from $- 6 x$ .

$x^{2} - 14 x = - 40$

Step 3.2.4

Add $40$ to both sides of the equation.

$x^{2} - 14 x + 40 = 0$

Step 3.2.5

Factor $x^{2} - 14 x + 40$ using the AC method.

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Step 3.2.5.1

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 $40$ and whose sum is $- 14$ .

$- 10, - 4$

Step 3.2.5.2

Write the factored form using these integers.

$(x - 10) (x - 4) = 0$

Step 3.2.6

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

$x - 10 = 0$

$x - 4 = 0$

Step 3.2.7

Set $x - 10$ equal to $0$ and solve for $x$ .

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Step 3.2.7.1

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 3.2.7.2

Add $10$ to both sides of the equation.

$x = 10$

Step 3.2.8

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

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Step 3.2.8.1

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

$x - 4 = 0$

Step 3.2.8.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.2.9

The final solution is all the values that make $(x - 10) (x - 4) = 0$ true.

$x = 10, 4$

Step 4

Exclude the solutions that do not make $\log (x - 6) - \log (8 x - 40) = \log (\frac{1}{x})$ true.

$x = 10$