Precalculus Examples

$2 \log_{3} (27) - \frac{1}{2} \cdot \log_{3} (9) + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $3$ of $27$ is $3$ .

$2 \cdot 3 - \frac{1}{2} \cdot \log_{3} (9) + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.2

Multiply $2$ by $3$ .

$6 - \frac{1}{2} \cdot \log_{3} (9) + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.3

Logarithm base $3$ of $9$ is $2$ .

$6 - \frac{1}{2} \cdot 2 + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$6 + \frac{- 1}{2} \cdot 2 + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.4.2

Cancel the common factor.

$6 + \frac{- 1}{2} \cdot 2 + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.4.3

Rewrite the expression.

$6 - 1 + \frac{1}{3} \cdot \log_{3} (27^{3}) - \log_{3} (x)$

Step 1.5

Raise $27$ to the power of $3$ .

$6 - 1 + \frac{1}{3} \cdot \log_{3} (19683) - \log_{3} (x)$

Step 1.6

Logarithm base $3$ of $19683$ is $9$ .

$6 - 1 + \frac{1}{3} \cdot 9 - \log_{3} (x)$

Step 1.7

Cancel the common factor of $3$ .

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Step 1.7.1

Factor $3$ out of $9$ .

$6 - 1 + \frac{1}{3} \cdot (3 (3)) - \log_{3} (x)$

Step 1.7.2

Cancel the common factor.

$6 - 1 + \frac{1}{3} \cdot (3 \cdot 3) - \log_{3} (x)$

Step 1.7.3

Rewrite the expression.

$6 - 1 + 3 - \log_{3} (x)$

Step 2

Simplify by adding and subtracting.

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Step 2.1

Subtract $1$ from $6$ .

$5 + 3 - \log_{3} (x)$

Step 2.2

Add $5$ and $3$ .

$8 - \log_{3} (x)$