Precalculus Examples

$\log_{3} (s + 45) - \log_{3} (s + 5) = \log_{3} (s)$

Step 1

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

$\log_{3} (\frac{s + 45}{s + 5}) = \log_{3} (s)$

Step 2

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

$\frac{s + 45}{s + 5} = s$

Step 3

Solve for $s$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$s + 5, 1$

Step 3.1.2

Remove parentheses.

$s + 5, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$s + 5$

Step 3.2

Multiply each term in $\frac{s + 45}{s + 5} = s$ by $s + 5$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{s + 45}{s + 5} = s$ by $s + 5$ .

$\frac{s + 45}{s + 5} (s + 5) = s (s + 5)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $s + 5$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{s + 45}{s + 5} (s + 5) = s (s + 5)$

Step 3.2.2.1.2

Rewrite the expression.

$s + 45 = s (s + 5)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Apply the distributive property.

$s + 45 = s \cdot s + s \cdot 5$

Step 3.2.3.2

Simplify the expression.

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Step 3.2.3.2.1

Multiply $s$ by $s$ .

$s + 45 = s^{2} + s \cdot 5$

Step 3.2.3.2.2

Move $5$ to the left of $s$ .

$s + 45 = s^{2} + 5 s$

Step 3.3

Solve the equation.

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Step 3.3.1

Since $s$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$s^{2} + 5 s = s + 45$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $s$ from both sides of the equation.

$s^{2} + 5 s - s = 45$

Step 3.3.2.2

Subtract $s$ from $5 s$ .

$s^{2} + 4 s = 45$

Step 3.3.3

Subtract $45$ from both sides of the equation.

$s^{2} + 4 s - 45 = 0$

Step 3.3.4

Factor $s^{2} + 4 s - 45$ using the AC method.

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Step 3.3.4.1

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 $- 45$ and whose sum is $4$ .

$- 5, 9$

Step 3.3.4.2

Write the factored form using these integers.

$(s - 5) (s + 9) = 0$

Step 3.3.5

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

$s - 5 = 0$

$s + 9 = 0$

Step 3.3.6

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

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Step 3.3.6.1

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

$s - 5 = 0$

Step 3.3.6.2

Add $5$ to both sides of the equation.

$s = 5$

Step 3.3.7

Set $s + 9$ equal to $0$ and solve for $s$ .

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Step 3.3.7.1

Set $s + 9$ equal to $0$ .

$s + 9 = 0$

Step 3.3.7.2

Subtract $9$ from both sides of the equation.

$s = - 9$

Step 3.3.8

The final solution is all the values that make $(s - 5) (s + 9) = 0$ true.

$s = 5, - 9$

Step 4

Exclude the solutions that do not make $\log_{3} (s + 45) - \log_{3} (s + 5) = \log_{3} (s)$ true.

$s = 5$