Precalculus Examples

$\frac{2}{3} \cdot \ln (27) + \ln (20) - 3 \ln (8)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{2}{3} \ln (27)$ by moving $\frac{2}{3}$ inside the logarithm.

$\ln (27^{\frac{2}{3}}) + \ln (20) - 3 \ln (8)$

Step 1.2

Rewrite $27$ as $3^{3}$ .

$\ln ({(3^{3})}^{\frac{2}{3}}) + \ln (20) - 3 \ln (8)$

Step 1.3

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\ln (3^{3 (\frac{2}{3})}) + \ln (20) - 3 \ln (8)$

Step 1.4

Cancel the common factor of $3$ .

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Step 1.4.1

Cancel the common factor.

$\ln (3^{3 (\frac{2}{3})}) + \ln (20) - 3 \ln (8)$

Step 1.4.2

Rewrite the expression.

$\ln (3^{2}) + \ln (20) - 3 \ln (8)$

Step 1.5

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

$\ln (9) + \ln (20) - 3 \ln (8)$

Step 1.6

Simplify $- 3 \ln (8)$ by moving $3$ inside the logarithm.

$\ln (9) + \ln (20) - \ln (8^{3})$

Step 1.7

Raise $8$ to the power of $3$ .

$\ln (9) + \ln (20) - \ln (512)$

Step 2

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

$\ln (9 \cdot 20) - \ln (512)$

Step 3

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

$\ln (\frac{9 \cdot 20}{512})$

Step 4

Cancel the common factor of $20$ and $512$ .

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Step 4.1

Factor $4$ out of $9 \cdot 20$ .

$\ln (\frac{4 (9 \cdot 5)}{512})$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $4$ out of $512$ .

$\ln (\frac{4 (9 \cdot 5)}{4 (128)})$

Step 4.2.2

Cancel the common factor.

$\ln (\frac{4 (9 \cdot 5)}{4 \cdot 128})$

Step 4.2.3

Rewrite the expression.

$\ln (\frac{9 \cdot 5}{128})$

Step 5

Multiply $9$ by $5$ .

$\ln (\frac{45}{128})$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\ln (\frac{45}{128})$

Decimal Form:

$- 1.04536777 \dots$