Precalculus Examples

$\log_{2} (2) \cdot \log_{2} (4) \cdot \log_{2} (8) \cdot \log_{2} (2^{n})$

Step 1

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

$1 \cdot \log_{2} (4) \cdot \log_{2} (8) \cdot \log_{2} (2^{n})$

Step 2

Multiply $\log_{2} (4)$ by $1$ .

$\log_{2} (4) \cdot \log_{2} (8) \cdot \log_{2} (2^{n})$

Step 3

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

$2 \cdot \log_{2} (8) \cdot \log_{2} (2^{n})$

Step 4

Logarithm base $2$ of $8$ is $3$ .

$2 \cdot 3 \cdot \log_{2} (2^{n})$

Step 5

Multiply $2$ by $3$ .

$6 \cdot \log_{2} (2^{n})$

Step 6

Use logarithm rules to move $n$ out of the exponent.

$6 \cdot (n \log_{2} (2))$

Step 7

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

$6 \cdot (n \cdot 1)$

Step 8

Multiply $n$ by $1$ .

$6 n$