Precalculus Examples

$(- \frac{\ln (3.3)}{12}) \cdot (\frac{1}{(1 + 0.0833 \cdot 0.04) \cdot (\ln ({(1 + 0.0833 \cdot 0.04)}^{2}))})$

Step 1

Rewrite $\frac{\ln (3.3)}{12}$ as $\frac{1}{12} \ln (3.3)$ .

$- (\frac{1}{12} \ln (3.3)) \cdot \frac{1}{(1 + 0.0833 \cdot 0.04) \cdot (\ln ({(1 + 0.0833 \cdot 0.04)}^{2}))}$

Step 2

Simplify $\frac{1}{12} \ln (3.3)$ by moving $\frac{1}{12}$ inside the logarithm.

$- \ln ({3.3}^{\frac{1}{12}}) \cdot \frac{1}{(1 + 0.0833 \cdot 0.04) \cdot (\ln ({(1 + 0.0833 \cdot 0.04)}^{2}))}$

Step 3

Raise $3.3$ to the power of $\frac{1}{12}$ .

$- \ln (1.10461133) \cdot \frac{1}{(1 + 0.0833 \cdot 0.04) \cdot (\ln ({(1 + 0.0833 \cdot 0.04)}^{2}))}$

Step 4

Simplify the denominator.

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Step 4.1

Multiply $0.0833$ by $0.04$ .

$- \ln (1.10461133) \cdot \frac{1}{(1 + 0.003332) \ln ({(1 + 0.0833 \cdot 0.04)}^{2})}$

Step 4.2

Add $1$ and $0.003332$ .

$- \ln (1.10461133) \cdot \frac{1}{1.003332 \ln ({(1 + 0.0833 \cdot 0.04)}^{2})}$

Step 4.3

Multiply $0.0833$ by $0.04$ .

$- \ln (1.10461133) \cdot \frac{1}{1.003332 \ln ({(1 + 0.003332)}^{2})}$

Step 4.4

Add $1$ and $0.003332$ .

$- \ln (1.10461133) \cdot \frac{1}{1.003332 \ln ({1.003332}^{2})}$

Step 4.5

Raise $1.003332$ to the power of $2$ .

$- \ln (1.10461133) \cdot \frac{1}{1.003332 \ln (1.0066751)}$

Step 5

Simplify $1.003332 \ln (1.0066751)$ by moving $1.003332$ inside the logarithm.

$- \ln (1.10461133) \cdot \frac{1}{\ln ({1.0066751}^{1.003332})}$

Step 6

Raise $1.0066751$ to the power of $1.003332$ .

$- \ln (1.10461133) \cdot \frac{1}{\ln (1.00669741)}$

Step 7

Combine $\frac{1}{\ln (1.00669741)}$ and $\ln (1.10461133)$ .

$- \frac{\ln (1.10461133)}{\ln (1.00669741)}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$- \frac{\ln (1.10461133)}{\ln (1.00669741)}$

Decimal Form:

$- 14.90519833 \dots$