Precalculus Examples

$3 \log_{2} (4) + 4 \log_{2} (\frac{1}{16}) - 2 \log_{2} (2) + \log_{2} (1)$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $2$ of $4$ is $2$ .

$3 \cdot 2 + 4 \log_{2} (\frac{1}{16}) - 2 \log_{2} (2) + \log_{2} (1)$

Step 1.2

Multiply $3$ by $2$ .

$6 + 4 \log_{2} (\frac{1}{16}) - 2 \log_{2} (2) + \log_{2} (1)$

Step 1.3

Logarithm base $2$ of $\frac{1}{16}$ is $- 4$ .

$6 + 4 \cdot - 4 - 2 \log_{2} (2) + \log_{2} (1)$

Step 1.4

Multiply $4$ by $- 4$ .

$6 - 16 - 2 \log_{2} (2) + \log_{2} (1)$

Step 1.5

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

$6 - 16 - 2 \cdot 1 + \log_{2} (1)$

Step 1.6

Multiply $- 2$ by $1$ .

$6 - 16 - 2 + \log_{2} (1)$

Step 1.7

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

$6 - 16 - 2 + 0$

Step 2

Simplify by adding and subtracting.

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Step 2.1

Subtract $16$ from $6$ .

$- 10 - 2 + 0$

Step 2.2

Subtract $2$ from $- 10$ .

$- 12 + 0$

Step 2.3

Add $- 12$ and $0$ .

$- 12$