Precalculus Examples

$3 \log_{2} (3 - \frac{1}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1

Simplify each term.

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Step 1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 \log_{2} (3 \cdot \frac{2}{2} - \frac{1}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.2

Combine $3$ and $\frac{2}{2}$ .

$3 \log_{2} (\frac{3 \cdot 2}{2} - \frac{1}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.3

Combine the numerators over the common denominator.

$3 \log_{2} (\frac{3 \cdot 2 - 1}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.4

Simplify the numerator.

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Step 1.4.1

Multiply $3$ by $2$ .

$3 \log_{2} (\frac{6 - 1}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.4.2

Subtract $1$ from $6$ .

$3 \log_{2} (\frac{5}{2}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.5

Rewrite $\frac{5}{2}$ as $5 \cdot 2^{- 1}$ .

$3 \log_{2} (5 \cdot 2^{- 1}) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.6

Rewrite $\log_{2} (5 \cdot 2^{- 1})$ as $\log_{2} (5) + \log_{2} (2^{- 1})$ .

$3 (\log_{2} (5) + \log_{2} (2^{- 1})) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.7

Use logarithm rules to move $- 1$ out of the exponent.

$3 (\log_{2} (5) - \log_{2} (2)) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.8

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

$3 (\log_{2} (5) - 1 \cdot 1) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.9

Multiply $- 1$ by $1$ .

$3 (\log_{2} (5) - 1) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.10

Apply the distributive property.

$(3 \log_{2} (5) + 3 \cdot - 1) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.11

Multiply $3$ by $- 1$ .

$(3 \log_{2} (5) - 3) \log_{2} (9) - 2 \log_{2} (6)$

Step 1.12

Rewrite $\log_{2} (9)$ as $\log_{2} (3^{2})$ .

$(3 \log_{2} (5) - 3) \log_{2} (3^{2}) - 2 \log_{2} (6)$

Step 1.13

Expand $\log_{2} (3^{2})$ by moving $2$ outside the logarithm.

$(3 \log_{2} (5) - 3) (2 \log_{2} (3)) - 2 \log_{2} (6)$

Step 1.14

Apply the distributive property.

$3 \log_{2} (5) (2 \log_{2} (3)) - 3 (2 \log_{2} (3)) - 2 \log_{2} (6)$

Step 1.15

Multiply $2$ by $3$ .

$6 \log_{2} (5) \log_{2} (3) - 3 (2 \log_{2} (3)) - 2 \log_{2} (6)$

Step 1.16

Multiply $2$ by $- 3$ .

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 \log_{2} (6)$

Step 1.17

Rewrite $\log_{2} (6)$ as $\log_{2} (2 (3))$ .

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 \log_{2} (2 (3))$

Step 1.18

Rewrite $\log_{2} (2 (3))$ as $\log_{2} (2) + \log_{2} (3)$ .

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 (\log_{2} (2) + \log_{2} (3))$

Step 1.19

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

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 (1 + \log_{2} (3))$

Step 1.20

Apply the distributive property.

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 \cdot 1 - 2 \log_{2} (3)$

Step 1.21

Multiply $- 2$ by $1$ .

$6 \log_{2} (5) \log_{2} (3) - 6 \log_{2} (3) - 2 - 2 \log_{2} (3)$

Step 2

Subtract $2 \log_{2} (3)$ from $- 6 \log_{2} (3)$ .

$6 \log_{2} (5) \log_{2} (3) - 2 - 8 \log_{2} (3)$

Step 3

The result can be shown in multiple forms.

Exact Form:

$6 \log_{2} (5) \log_{2} (3) - 2 - 8 \log_{2} (3)$

Decimal Form:

$7.40131375 \dots$