Precalculus Examples

$\log_{2} (64) + \log_{3} (27) \cdot \log_{4} (0.00390625)$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $2$ of $64$ is $6$ .

$6 + \log_{3} (27) \cdot \log_{4} (0.00390625)$

Step 1.2

Logarithm base $3$ of $27$ is $3$ .

$6 + 3 \cdot \log_{4} (0.00390625)$

Step 1.3

Log base $4$ of $0.00390625$ is approximately $- 4$ .

$6 + 3 \cdot - 4$

Step 1.4

Multiply $3$ by $- 4$ .

$6 - 12$

Step 2

Subtract $12$ from $6$ .

$- 6$