Calculus Examples

$\frac{\ln (\frac{4}{9})}{- 0.5}$

Step 1

Rewrite $\ln (\frac{4}{9})$ as $\ln (4) - \ln (9)$ .

$\frac{\ln (4) - \ln (9)}{- 0.5}$

Step 2

Rewrite $\ln (9)$ as $\ln (3^{2})$ .

$\frac{\ln (4) - \ln (3^{2})}{- 0.5}$

Step 3

Expand $\ln (3^{2})$ by moving $2$ outside the logarithm.

$\frac{\ln (4) - (2 \ln (3))}{- 0.5}$

Step 4

Multiply $2$ by $- 1$ .

$\frac{\ln (4) - 2 \ln (3)}{- 0.5}$

Step 5

Simplify the numerator.

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Step 5.1

Rewrite $\ln (4)$ as $\ln (2^{2})$ .

$\frac{\ln (2^{2}) - 2 \ln (3)}{- 0.5}$

Step 5.2

Expand $\ln (2^{2})$ by moving $2$ outside the logarithm.

$\frac{2 \ln (2) - 2 \ln (3)}{- 0.5}$

Step 6

Move the negative in front of the fraction.

$- \frac{2 \ln (2) - 2 \ln (3)}{0.5}$

Step 7

Replace $e$ with an approximation.

$- \frac{2 \log_{2.71828182} (2) - 2 \ln (3)}{0.5}$

Step 8

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

$- \frac{2 \cdot 0.69314718 - 2 \ln (3)}{0.5}$

Step 9

Multiply $2$ by $0.69314718$ .

$- \frac{1.38629436 - 2 \ln (3)}{0.5}$

Step 10

Replace $e$ with an approximation.

$- \frac{1.38629436 - 2 \log_{2.71828182} (3)}{0.5}$

Step 11

Log base $2.71828182$ of $3$ is approximately $1.09861228$ .

$- \frac{1.38629436 - 2 \cdot 1.09861228}{0.5}$

Step 12

Multiply $- 2$ by $1.09861228$ .

$- \frac{1.38629436 - 2.19722457}{0.5}$

Step 13

Subtract $2.19722457$ from $1.38629436$ .

$- \frac{- 0.81093021}{0.5}$

Step 14

Divide $- 0.81093021$ by $0.5$ .

$- - 1.62186043$

Step 15

Multiply $- 1$ by $- 1.62186043$ .

$1.62186043$