Precalculus Examples

$\log_{2} (2) \cdot \log_{2} (4) \cdot \log_{2} (6) \cdot \log_{2} (2^{n})$

Step 1

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

$1 \cdot \log_{2} (4) \cdot \log_{2} (6) \cdot \log_{2} (2^{n})$

Step 2

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

$\log_{2} (4) \cdot \log_{2} (6) \cdot \log_{2} (2^{n})$

Step 3

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

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

Step 4

Simplify $2 \log_{2} (6)$ by moving $2$ inside the logarithm.

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

Step 5

Raise $6$ to the power of $2$ .

$\log_{2} (36) \cdot \log_{2} (2^{n})$

Step 6

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

$\log_{2} (36) \cdot (n \log_{2} (2))$

Step 7

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

$\log_{2} (36) \cdot (n \cdot 1)$

Step 8

Multiply $n$ by $1$ .

$\log_{2} (36) \cdot n$

Step 9

Reorder factors in $\log_{2} (36) n$ .

$n \log_{2} (36)$