Precalculus Examples

$\frac{\ln (2)}{- \frac{1}{20} \cdot \ln (\frac{3.1}{10.6})}$

Step 1

Simplify the denominator.

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Step 1.1

Reorder $\ln (\frac{3.1}{10.6})$ and $\frac{1}{20}$ .

$\frac{\ln (2)}{- (\frac{1}{20} \ln (\frac{3.1}{10.6}))}$

Step 1.2

Simplify $\frac{1}{20} \ln (\frac{3.1}{10.6})$ by moving $\frac{1}{20}$ inside the logarithm.

$\frac{\ln (2)}{- \ln ({(\frac{3.1}{10.6})}^{\frac{1}{20}})}$

Step 2

Simplify the denominator.

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Step 2.1

Divide $3.1$ by $10.6$ .

$\frac{\ln (2)}{- \ln ({0.29245283}^{\frac{1}{20}})}$

Step 2.2

Raise $0.29245283$ to the power of $\frac{1}{20}$ .

$\frac{\ln (2)}{- \ln (0.94037871)}$

Step 3

Move the negative in front of the fraction.

$- \frac{\ln (2)}{\ln (0.94037871)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- \frac{\ln (2)}{\ln (0.94037871)}$

Decimal Form:

$11.27571052 \dots$