Precalculus Examples

$\log_{3} (60) - \log_{3} (12) - \log_{3} (35)$

Step 1

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

$\log_{3} (\frac{60}{12}) - \log_{3} (35)$

Step 2

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

$\log_{3} (\frac{\frac{60}{12}}{35})$

Step 3

Cancel the common factor of $60$ and $12$ .

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Step 3.1

Factor $12$ out of $60$ .

$\log_{3} (\frac{\frac{12 \cdot 5}{12}}{35})$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $12$ out of $12$ .

$\log_{3} (\frac{\frac{12 \cdot 5}{12 (1)}}{35})$

Step 3.2.2

Cancel the common factor.

$\log_{3} (\frac{\frac{12 \cdot 5}{12 \cdot 1}}{35})$

Step 3.2.3

Rewrite the expression.

$\log_{3} (\frac{\frac{5}{1}}{35})$

Step 3.2.4

Divide $5$ by $1$ .

$\log_{3} (\frac{5}{35})$

Step 4

Cancel the common factor of $5$ and $35$ .

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Step 4.1

Factor $5$ out of $5$ .

$\log_{3} (\frac{5 (1)}{35})$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $5$ out of $35$ .

$\log_{3} (\frac{5 \cdot 1}{5 \cdot 7})$

Step 4.2.2

Cancel the common factor.

$\log_{3} (\frac{5 \cdot 1}{5 \cdot 7})$

Step 4.2.3

Rewrite the expression.

$\log_{3} (\frac{1}{7})$

Step 5

Rewrite $\log_{3} (\frac{1}{7})$ using the change of base formula.

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Step 5.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 5.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (\frac{1}{7})}{\log (3)}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (\frac{1}{7})}{\log (3)}$

Decimal Form:

$- 1.77124374 \dots$