Precalculus Examples

$2 \log (5) + 3 \log (4) - 4 \log (2)$

Step 1

Simplify each term.

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Step 1.1

Simplify $2 \log (5)$ by moving $2$ inside the logarithm.

$\log (5^{2}) + 3 \log (4) - 4 \log (2)$

Step 1.2

Raise $5$ to the power of $2$ .

$\log (25) + 3 \log (4) - 4 \log (2)$

Step 1.3

Simplify $3 \log (4)$ by moving $3$ inside the logarithm.

$\log (25) + \log (4^{3}) - 4 \log (2)$

Step 1.4

Raise $4$ to the power of $3$ .

$\log (25) + \log (64) - 4 \log (2)$

Step 1.5

Simplify $- 4 \log (2)$ by moving $4$ inside the logarithm.

$\log (25) + \log (64) - \log (2^{4})$

Step 1.6

Raise $2$ to the power of $4$ .

$\log (25) + \log (64) - \log (16)$

Step 2

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

$\log (25 \cdot 64) - \log (16)$

Step 3

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

$\log (\frac{25 \cdot 64}{16})$

Step 4

Cancel the common factor of $64$ and $16$ .

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Step 4.1

Factor $16$ out of $25 \cdot 64$ .

$\log (\frac{16 (25 \cdot 4)}{16})$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $16$ out of $16$ .

$\log (\frac{16 (25 \cdot 4)}{16 (1)})$

Step 4.2.2

Cancel the common factor.

$\log (\frac{16 (25 \cdot 4)}{16 \cdot 1})$

Step 4.2.3

Rewrite the expression.

$\log (\frac{25 \cdot 4}{1})$

Step 4.2.4

Divide $25 \cdot 4$ by $1$ .

$\log (25 \cdot 4)$

Step 5

Simplify the expression.

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Step 5.1

Multiply $25$ by $4$ .

$\log (100)$

Step 5.2

Logarithm base $10$ of $100$ is $2$ .

$2$