Precalculus Examples

$\frac{1}{2} \cdot \ln (9) + \frac{1}{3} \cdot \ln (64)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{2} \ln (9)$ by moving $\frac{1}{2}$ inside the logarithm.

$\ln (9^{\frac{1}{2}}) + \frac{1}{3} \cdot \ln (64)$

Step 1.2

Rewrite $9$ as $3^{2}$ .

$\ln ({(3^{2})}^{\frac{1}{2}}) + \frac{1}{3} \cdot \ln (64)$

Step 1.3

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

$\ln (3^{2 (\frac{1}{2})}) + \frac{1}{3} \cdot \ln (64)$

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Cancel the common factor.

$\ln (3^{2 (\frac{1}{2})}) + \frac{1}{3} \cdot \ln (64)$

Step 1.4.2

Rewrite the expression.

$\ln (3^{1}) + \frac{1}{3} \cdot \ln (64)$

Step 1.5

Evaluate the exponent.

$\ln (3) + \frac{1}{3} \cdot \ln (64)$

Step 1.6

Simplify $\frac{1}{3} \ln (64)$ by moving $\frac{1}{3}$ inside the logarithm.

$\ln (3) + \ln (64^{\frac{1}{3}})$

Step 1.7

Rewrite $64$ as $4^{3}$ .

$\ln (3) + \ln ({(4^{3})}^{\frac{1}{3}})$

Step 1.8

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

$\ln (3) + \ln (4^{3 (\frac{1}{3})})$

Step 1.9

Cancel the common factor of $3$ .

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Step 1.9.1

Cancel the common factor.

$\ln (3) + \ln (4^{3 (\frac{1}{3})})$

Step 1.9.2

Rewrite the expression.

$\ln (3) + \ln (4^{1})$

Step 1.10

Evaluate the exponent.

$\ln (3) + \ln (4)$

Step 2

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

$\ln (3 \cdot 4)$

Step 3

Multiply $3$ by $4$ .

$\ln (12)$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\ln (12)$

Decimal Form:

$2.48490664 \dots$