Precalculus Examples

$\frac{\ln (2)}{r} - \frac{\ln (2)}{0.023}$

Step 1

Simplify each term.

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Step 1.1

Replace $e$ with an approximation.

$\frac{\ln (2)}{r} - \frac{\log_{2.71828182} (2)}{0.023}$

Step 1.2

Log base $2.71828182$ of $2$ is approximately $0.69314718$ .

$\frac{\ln (2)}{r} - \frac{0.69314718}{0.023}$

Step 1.3

Divide $0.69314718$ by $0.023$ .

$\frac{\ln (2)}{r} - 1 \cdot 30.13683393$

Step 1.4

Multiply $- 1$ by $30.13683393$ .

$\frac{\ln (2)}{r} - 30.13683393$

Step 2

To write $- 30.13683393$ as a fraction with a common denominator, multiply by $\frac{r}{r}$ .

$\frac{\ln (2)}{r} - 30.13683393 \cdot \frac{r}{r}$

Step 3

Combine $- 30.13683393$ and $\frac{r}{r}$ .

$\frac{\ln (2)}{r} + \frac{- 30.13683393 r}{r}$

Step 4

Combine the numerators over the common denominator.

$\frac{\ln (2) - 30.13683393 r}{r}$