Precalculus Examples

$\log_{3} (a - 2) + \log_{3} (a - 2) = 2$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{3} ((a - 2) (a - 2)) = 2$

Step 1.2

Expand $(a - 2) (a - 2)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{3} (a (a - 2) - 2 (a - 2)) = 2$

Step 1.2.2

Apply the distributive property.

$\log_{3} (a \cdot a + a \cdot - 2 - 2 (a - 2)) = 2$

Step 1.2.3

Apply the distributive property.

$\log_{3} (a \cdot a + a \cdot - 2 - 2 a - 2 \cdot - 2) = 2$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $a$ by $a$ .

$\log_{3} (a^{2} + a \cdot - 2 - 2 a - 2 \cdot - 2) = 2$

Step 1.3.1.2

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

$\log_{3} (a^{2} - 2 \cdot a - 2 a - 2 \cdot - 2) = 2$

Step 1.3.1.3

Multiply $- 2$ by $- 2$ .

$\log_{3} (a^{2} - 2 a - 2 a + 4) = 2$

Step 1.3.2

Subtract $2 a$ from $- 2 a$ .

$\log_{3} (a^{2} - 4 a + 4) = 2$

Step 2

Rewrite $\log_{3} (a^{2} - 4 a + 4) = 2$ 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 $b^{y} = x$ .

$3^{2} = a^{2} - 4 a + 4$

Step 3

Solve for $a$ .

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Step 3.1

Rewrite the equation as $a^{2} - 4 a + 4 = 3^{2}$ .

$a^{2} - 4 a + 4 = 3^{2}$

Step 3.2

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

$a^{2} - 4 a + 4 = 9$

Step 3.3

Subtract $9$ from both sides of the equation.

$a^{2} - 4 a + 4 - 9 = 0$

Step 3.4

Subtract $9$ from $4$ .

$a^{2} - 4 a - 5 = 0$

Step 3.5

Factor $a^{2} - 4 a - 5$ using the AC method.

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Step 3.5.1

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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 3.5.2

Write the factored form using these integers.

$(a - 5) (a + 1) = 0$

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 $0$ .

$a - 5 = 0$

$a + 1 = 0$

Step 3.7

Set $a - 5$ equal to $0$ and solve for $a$ .

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Step 3.7.1

Set $a - 5$ equal to $0$ .

$a - 5 = 0$

Step 3.7.2

Add $5$ to both sides of the equation.

$a = 5$

Step 3.8

Set $a + 1$ equal to $0$ and solve for $a$ .

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Step 3.8.1

Set $a + 1$ equal to $0$ .

$a + 1 = 0$

Step 3.8.2

Subtract $1$ from both sides of the equation.

$a = - 1$

Step 3.9

The final solution is all the values that make $(a - 5) (a + 1) = 0$ true.

$a = 5, - 1$

Step 4

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

$a = 5$