Precalculus Examples

$\log_{9} (3) + \log_{9} (\frac{1}{27}) + \log_{2} (1) - (\log_{3} (\sqrt{27}))$

Step 1

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

$\log_{9} (3 (\frac{1}{27})) + \log_{2} (1) - (\log_{3} (\sqrt{27}))$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of $3$ .

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Step 2.1.1

Factor $3$ out of $27$ .

$\log_{9} (3 \frac{1}{3 (9)}) + \log_{2} (1) - (\log_{3} (\sqrt{27}))$

Step 2.1.2

Cancel the common factor.

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

Step 2.1.3

Rewrite the expression.

$\log_{9} (\frac{1}{9}) + \log_{2} (1) - (\log_{3} (\sqrt{27}))$

Step 2.2

Logarithm base $9$ of $\frac{1}{9}$ is $- 1$ .

$- 1 + \log_{2} (1) - (\log_{3} (\sqrt{27}))$

Step 2.3

Logarithm base $2$ of $1$ is $0$ .

$- 1 + 0 - (\log_{3} (\sqrt{27}))$

Step 2.4

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 2.4.1

Factor $9$ out of $27$ .

$- 1 + 0 - \log_{3} (\sqrt{9 (3)})$

Step 2.4.2

Rewrite $9$ as $3^{2}$ .

$- 1 + 0 - \log_{3} (\sqrt{3^{2} \cdot 3})$

Step 2.5

Pull terms out from under the radical.

$- 1 + 0 - \log_{3} (3 \sqrt{3})$

Step 3

Add $- 1$ and $0$ .

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$- 2.5$