Precalculus Examples

$\log_{2} (x - 6) = 5 - \log_{2} (2 x)$

Step 1

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

$\log_{2} (x - 6) + \log_{2} (2 x) = 5$

Step 2

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

$\log_{2} ((x - 6) (2 x)) = 5$

Step 3

Apply the distributive property.

$\log_{2} (x (2 x) - 6 (2 x)) = 5$

Step 4

Simplify the expression.

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Step 4.1

Rewrite using the commutative property of multiplication.

$\log_{2} (2 x \cdot x - 6 (2 x)) = 5$

Step 4.2

Multiply $2$ by $- 6$ .

$\log_{2} (2 x \cdot x - 12 x) = 5$

Step 5

Multiply $x$ by $x$ by adding the exponents.

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Step 5.1

Move $x$ .

$\log_{2} (2 (x \cdot x) - 12 x) = 5$

Step 5.2

Multiply $x$ by $x$ .

$\log_{2} (2 x^{2} - 12 x) = 5$

Step 6

Rewrite $\log_{2} (2 x^{2} - 12 x) = 5$ 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 $b^{y} = x$ .

$2^{5} = 2 x^{2} - 12 x$

Step 7

Solve for $x$ .

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Step 7.1

Rewrite the equation as $2 x^{2} - 12 x = 2^{5}$ .

$2 x^{2} - 12 x = 2^{5}$

Step 7.2

Raise $2$ to the power of $5$ .

$2 x^{2} - 12 x = 32$

Step 7.3

Subtract $32$ from both sides of the equation.

$2 x^{2} - 12 x - 32 = 0$

Step 7.4

Factor the left side of the equation.

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Step 7.4.1

Factor $2$ out of $2 x^{2} - 12 x - 32$ .

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Step 7.4.1.1

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

$2 (x^{2}) - 12 x - 32 = 0$

Step 7.4.1.2

Factor $2$ out of $- 12 x$ .

$2 (x^{2}) + 2 (- 6 x) - 32 = 0$

Step 7.4.1.3

Factor $2$ out of $- 32$ .

$2 x^{2} + 2 (- 6 x) + 2 \cdot - 16 = 0$

Step 7.4.1.4

Factor $2$ out of $2 x^{2} + 2 (- 6 x)$ .

$2 (x^{2} - 6 x) + 2 \cdot - 16 = 0$

Step 7.4.1.5

Factor $2$ out of $2 (x^{2} - 6 x) + 2 \cdot - 16$ .

$2 (x^{2} - 6 x - 16) = 0$

Step 7.4.2

Factor.

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Step 7.4.2.1

Factor $x^{2} - 6 x - 16$ using the AC method.

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Step 7.4.2.1.1

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 $- 16$ and whose sum is $- 6$ .

$- 8, 2$

Step 7.4.2.1.2

Write the factored form using these integers.

$2 ((x - 8) (x + 2)) = 0$

Step 7.4.2.2

Remove unnecessary parentheses.

$2 (x - 8) (x + 2) = 0$

Step 7.5

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

$x - 8 = 0$

$x + 2 = 0$

Step 7.6

Set $x - 8$ equal to $0$ and solve for $x$ .

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Step 7.6.1

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 7.6.2

Add $8$ to both sides of the equation.

$x = 8$

Step 7.7

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

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Step 7.7.1

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

$x + 2 = 0$

Step 7.7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7.8

The final solution is all the values that make $2 (x - 8) (x + 2) = 0$ true.

$x = 8, - 2$

Step 8

Exclude the solutions that do not make $\log_{2} (x - 6) = 5 - \log_{2} (2 x)$ true.

$x = 8$