Precalculus Examples

$\log (40) + 3 \log (2) - \log (32)$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$3 \log (2) + \log (\frac{40}{32})$

Step 2

Simplify each term.

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Step 2.1

Simplify $3 \log (2)$ by moving $3$ inside the logarithm.

$\log (2^{3}) + \log (\frac{40}{32})$

Step 2.2

Raise $2$ to the power of $3$ .

$\log (8) + \log (\frac{40}{32})$

Step 2.3

Cancel the common factor of $40$ and $32$ .

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Step 2.3.1

Factor $8$ out of $40$ .

$\log (8) + \log (\frac{8 (5)}{32})$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor $8$ out of $32$ .

$\log (8) + \log (\frac{8 \cdot 5}{8 \cdot 4})$

Step 2.3.2.2

Cancel the common factor.

$\log (8) + \log (\frac{8 \cdot 5}{8 \cdot 4})$

Step 2.3.2.3

Rewrite the expression.

$\log (8) + \log (\frac{5}{4})$

Step 3

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (8 (\frac{5}{4}))$

Step 4

Cancel the common factor of $4$ .

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Step 4.1

Factor $4$ out of $8$ .

$\log (4 (2) \frac{5}{4})$

Step 4.2

Cancel the common factor.

$\log (4 \cdot 2 \frac{5}{4})$

Step 4.3

Rewrite the expression.

$\log (2 \cdot 5)$

Step 5

Multiply $2$ by $5$ .

$\log (10)$

Step 6

Logarithm base $10$ of $10$ is $1$ .

$1$