Precalculus Examples

$\log_{3} (72) + \log_{4} (6)$

Step 1

Simplify each term.

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Step 1.1

Rewrite $\log_{3} (72)$ as $\log_{3} (2^{3} \cdot 3^{2})$ .

$\log_{3} (2^{3} \cdot 3^{2}) + \log_{4} (6)$

Step 1.2

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

$\log_{3} (2^{3}) + \log_{3} (3^{2}) + \log_{4} (6)$

Step 1.3

Use logarithm rules to move $2$ out of the exponent.

$\log_{3} (2^{3}) + 2 \log_{3} (3) + \log_{4} (6)$

Step 1.4

Logarithm base $3$ of $3$ is $1$ .

$\log_{3} (2^{3}) + 2 \cdot 1 + \log_{4} (6)$

Step 1.5

Multiply $2$ by $1$ .

$\log_{3} (2^{3}) + 2 + \log_{4} (6)$

Step 1.6

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

$3 \log_{3} (2) + 2 + \log_{4} (6)$

Step 1.7

Rewrite $\log_{4} (6)$ as $\log_{4} (2 (3))$ .

$3 \log_{3} (2) + 2 + \log_{4} (2 (3))$

Step 1.8

Rewrite $\log_{4} (2 (3))$ as $\log_{4} (2) + \log_{4} (3)$ .

$3 \log_{3} (2) + 2 + \log_{4} (2) + \log_{4} (3)$

Step 1.9

Logarithm base $4$ of $2$ is $\frac{1}{2}$ .

$3 \log_{3} (2) + 2 + \frac{1}{2} + \log_{4} (3)$

Step 2

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

$3 \log_{3} (2) + 2 \cdot \frac{2}{2} + \frac{1}{2} + \log_{4} (3)$

Step 3

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

$3 \log_{3} (2) + \frac{2 \cdot 2}{2} + \frac{1}{2} + \log_{4} (3)$

Step 4

Combine the numerators over the common denominator.

$3 \log_{3} (2) + \frac{2 \cdot 2 + 1}{2} + \log_{4} (3)$

Step 5

Simplify the numerator.

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Step 5.1

Multiply $2$ by $2$ .

$3 \log_{3} (2) + \frac{4 + 1}{2} + \log_{4} (3)$

Step 5.2

Add $4$ and $1$ .

$3 \log_{3} (2) + \frac{5}{2} + \log_{4} (3)$