Precalculus Examples

$\log_{3} (x) - \log_{3} (6) + \log_{3} (2)$

Step 1

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

$\log_{3} (\frac{x}{6}) + \log_{3} (2)$

Step 2

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

$\log_{3} (\frac{x}{6} \cdot 2)$

Step 3

Cancel the common factor of $2$ .

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Step 3.1

Factor $2$ out of $6$ .

$\log_{3} (\frac{x}{2 (3)} \cdot 2)$

Step 3.2

Cancel the common factor.

$\log_{3} (\frac{x}{2 \cdot 3} \cdot 2)$

Step 3.3

Rewrite the expression.

$\log_{3} (\frac{x}{3})$

Step 4

Rewrite $\frac{x}{3}$ as $x \cdot 3^{- 1}$ .

$\log_{3} (x \cdot 3^{- 1})$

Step 5

Rewrite $\log_{3} (x \cdot 3^{- 1})$ as $\log_{3} (x) + \log_{3} (3^{- 1})$ .

$\log_{3} (x) + \log_{3} (3^{- 1})$

Step 6

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{3} (x) - \log_{3} (3)$

Step 7

Logarithm base $3$ of $3$ is $1$ .

$\log_{3} (x) - 1 \cdot 1$

Step 8

Multiply $- 1$ by $1$ .

$\log_{3} (x) - 1$

Step 9

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

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Step 9.1

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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} - 1$

Step 9.2

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

$\frac{\log (x)}{\log (3)} - 1$