Precalculus Examples

$\frac{1}{4} \cdot \log_{3} (x) + 3 \log_{3} (2 x) - \log_{3} (9)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{4} \log_{3} (x)$ by moving $\frac{1}{4}$ inside the logarithm.

$\log_{3} (x^{\frac{1}{4}}) + 3 \log_{3} (2 x) - \log_{3} (9)$

Step 1.2

Simplify $3 \log_{3} (2 x)$ by moving $3$ inside the logarithm.

$\log_{3} (x^{\frac{1}{4}}) + \log_{3} ({(2 x)}^{3}) - \log_{3} (9)$

Step 1.3

Apply the product rule to $2 x$ .

$\log_{3} (x^{\frac{1}{4}}) + \log_{3} (2^{3} x^{3}) - \log_{3} (9)$

Step 1.4

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

$\log_{3} (x^{\frac{1}{4}}) + \log_{3} (8 x^{3}) - \log_{3} (9)$

Step 1.5

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

$\log_{3} (x^{\frac{1}{4}}) + \log_{3} (8 x^{3}) - 1 \cdot 2$

Step 1.6

Multiply $- 1$ by $2$ .

$\log_{3} (x^{\frac{1}{4}}) + \log_{3} (8 x^{3}) - 2$

Step 2

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

$\log_{3} (x^{\frac{1}{4}} (8 x^{3})) - 2$

Step 3

Simplify each term.

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Step 3.1

Rewrite using the commutative property of multiplication.

$\log_{3} (8 x^{\frac{1}{4}} x^{3}) - 2$

Step 3.2

Multiply $x^{\frac{1}{4}}$ by $x^{3}$ by adding the exponents.

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Step 3.2.1

Move $x^{3}$ .

$\log_{3} (8 (x^{3} x^{\frac{1}{4}})) - 2$

Step 3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\log_{3} (8 x^{3 + \frac{1}{4}}) - 2$

Step 3.2.3

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\log_{3} (8 x^{3 \cdot \frac{4}{4} + \frac{1}{4}}) - 2$

Step 3.2.4

Combine $3$ and $\frac{4}{4}$ .

$\log_{3} (8 x^{\frac{3 \cdot 4}{4} + \frac{1}{4}}) - 2$

Step 3.2.5

Combine the numerators over the common denominator.

$\log_{3} (8 x^{\frac{3 \cdot 4 + 1}{4}}) - 2$

Step 3.2.6

Simplify the numerator.

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Step 3.2.6.1

Multiply $3$ by $4$ .

$\log_{3} (8 x^{\frac{12 + 1}{4}}) - 2$

Step 3.2.6.2

Add $12$ and $1$ .

$\log_{3} (8 x^{\frac{13}{4}}) - 2$