Precalculus Examples

$\log_{2} (x - 5) = 3 - \log_{2} (x - 7)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{2} (x - 5) + \log_{2} (x - 7) = 3$

Step 2

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

$\log_{2} ((x - 5) (x - 7)) = 3$

Step 3

Expand $(x - 5) (x - 7)$ using the FOIL Method.

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Step 3.1

Apply the distributive property.

$\log_{2} (x (x - 7) - 5 (x - 7)) = 3$

Step 3.2

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 7 - 5 (x - 7)) = 3$

Step 3.3

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 7 - 5 x - 5 \cdot - 7) = 3$

Step 4

Simplify and combine like terms.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $x$ by $x$ .

$\log_{2} (x^{2} + x \cdot - 7 - 5 x - 5 \cdot - 7) = 3$

Step 4.1.2

Move $- 7$ to the left of $x$ .

$\log_{2} (x^{2} - 7 \cdot x - 5 x - 5 \cdot - 7) = 3$

Step 4.1.3

Multiply $- 5$ by $- 7$ .

$\log_{2} (x^{2} - 7 x - 5 x + 35) = 3$

Step 4.2

Subtract $5 x$ from $- 7 x$ .

$\log_{2} (x^{2} - 12 x + 35) = 3$

Step 5

Rewrite $\log_{2} (x^{2} - 12 x + 35) = 3$ 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 $b^{y} = x$ .

$2^{3} = x^{2} - 12 x + 35$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $x^{2} - 12 x + 35 = 2^{3}$ .

$x^{2} - 12 x + 35 = 2^{3}$

Step 6.2

Raise $2$ to the power of $3$ .

$x^{2} - 12 x + 35 = 8$

Step 6.3

Subtract $8$ from both sides of the equation.

$x^{2} - 12 x + 35 - 8 = 0$

Step 6.4

Subtract $8$ from $35$ .

$x^{2} - 12 x + 27 = 0$

Step 6.5

Factor $x^{2} - 12 x + 27$ using the AC method.

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Step 6.5.1

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 $27$ and whose sum is $- 12$ .

$- 9, - 3$

Step 6.5.2

Write the factored form using these integers.

$(x - 9) (x - 3) = 0$

Step 6.6

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

$x - 9 = 0$

$x - 3 = 0$

Step 6.7

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

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Step 6.7.1

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

$x - 9 = 0$

Step 6.7.2

Add $9$ to both sides of the equation.

$x = 9$

Step 6.8

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.8.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.8.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6.9

The final solution is all the values that make $(x - 9) (x - 3) = 0$ true.

$x = 9, 3$

Step 7

Exclude the solutions that do not make $\log_{2} (x - 5) = 3 - \log_{2} (x - 7)$ true.

$x = 9$