Precalculus Examples

$\log_{6} (\frac{1}{36}) - \log_{2} (16)$

Step 1

Simplify each term.

Tap for more steps...

Step 1.1

Logarithm base $6$ of $\frac{1}{36}$ is $- 2$ .

$- 2 - \log_{2} (16)$

Step 1.2

Logarithm base $2$ of $16$ is $4$ .

$- 2 - 1 \cdot 4$

Step 1.3

Multiply $- 1$ by $4$ .

$- 2 - 4$

$- 2 - 4$

Step 2

Subtract $4$ from $- 2$ .

$- 6$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$