Precalculus Examples

$\log (2 x - 2) + \log (x - 6) = 2$

Step 1

Simplify the left side.

Tap for more steps...

Step 1.1

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

$\log ((2 x - 2) (x - 6)) = 2$

Step 1.2

Expand $(2 x - 2) (x - 6)$ using the FOIL Method.

Tap for more steps...

Step 1.2.1

Apply the distributive property.

$\log (2 x (x - 6) - 2 (x - 6)) = 2$

Step 1.2.2

Apply the distributive property.

$\log (2 x \cdot x + 2 x \cdot - 6 - 2 (x - 6)) = 2$

Step 1.2.3

Apply the distributive property.

$\log (2 x \cdot x + 2 x \cdot - 6 - 2 x - 2 \cdot - 6) = 2$

Step 1.3

Simplify and combine like terms.

Tap for more steps...

Step 1.3.1

Simplify each term.

Tap for more steps...

Step 1.3.1.1

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

Tap for more steps...

Step 1.3.1.1.1

Move $x$ .

$\log (2 (x \cdot x) + 2 x \cdot - 6 - 2 x - 2 \cdot - 6) = 2$

Step 1.3.1.1.2

Multiply $x$ by $x$ .

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

Step 1.3.1.2

Multiply $- 6$ by $2$ .

$\log (2 x^{2} - 12 x - 2 x - 2 \cdot - 6) = 2$

Step 1.3.1.3

Multiply $- 2$ by $- 6$ .

$\log (2 x^{2} - 12 x - 2 x + 12) = 2$

Step 1.3.2

Subtract $2 x$ from $- 12 x$ .

$\log (2 x^{2} - 14 x + 12) = 2$

Step 2

Rewrite $\log (2 x^{2} - 14 x + 12) = 2$ 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 $b^{y} = x$ .

$10^{2} = 2 x^{2} - 14 x + 12$

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

Rewrite the equation as $2 x^{2} - 14 x + 12 = 10^{2}$ .

$2 x^{2} - 14 x + 12 = 10^{2}$

Step 3.2

Raise $10$ to the power of $2$ .

$2 x^{2} - 14 x + 12 = 100$

Step 3.3

Subtract $100$ from both sides of the equation.

$2 x^{2} - 14 x + 12 - 100 = 0$

Step 3.4

Subtract $100$ from $12$ .

$2 x^{2} - 14 x - 88 = 0$

Step 3.5

Factor the left side of the equation.

Tap for more steps...

Step 3.5.1

Factor $2$ out of $2 x^{2} - 14 x - 88$ .

Tap for more steps...

Step 3.5.1.1

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

$2 (x^{2}) - 14 x - 88 = 0$

Step 3.5.1.2

Factor $2$ out of $- 14 x$ .

$2 (x^{2}) + 2 (- 7 x) - 88 = 0$

Step 3.5.1.3

Factor $2$ out of $- 88$ .

$2 x^{2} + 2 (- 7 x) + 2 \cdot - 44 = 0$

Step 3.5.1.4

Factor $2$ out of $2 x^{2} + 2 (- 7 x)$ .

$2 (x^{2} - 7 x) + 2 \cdot - 44 = 0$

Step 3.5.1.5

Factor $2$ out of $2 (x^{2} - 7 x) + 2 \cdot - 44$ .

$2 (x^{2} - 7 x - 44) = 0$

Step 3.5.2

Factor.

Tap for more steps...

Step 3.5.2.1

Factor $x^{2} - 7 x - 44$ using the AC method.

Tap for more steps...

Step 3.5.2.1.1

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 $- 44$ and whose sum is $- 7$ .

$- 11, 4$

Step 3.5.2.1.2

Write the factored form using these integers.

$2 ((x - 11) (x + 4)) = 0$

Step 3.5.2.2

Remove unnecessary parentheses.

$2 (x - 11) (x + 4) = 0$

Step 3.6

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

$x - 11 = 0$

$x + 4 = 0$

Step 3.7

Set $x - 11$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.7.1

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Step 3.7.2

Add $11$ to both sides of the equation.

$x = 11$

Step 3.8

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

Tap for more steps...

Step 3.8.1

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

$x + 4 = 0$

Step 3.8.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 3.9

The final solution is all the values that make $2 (x - 11) (x + 4) = 0$ true.

$x = 11, - 4$

Step 4

Exclude the solutions that do not make $\log (2 x - 2) + \log (x - 6) = 2$ true.

$x = 11$