Calculus Examples

$\ln (\frac{30}{142}) = - 0.03 t$

Step 1

Rewrite the equation as $- 0.03 t = \ln (\frac{30}{142})$ .

$- 0.03 t = \ln (\frac{30}{142})$

Step 2

Reduce the expression $\frac{30}{142}$ by cancelling the common factors.

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Step 2.1

Factor $2$ out of $30$ .

$- 0.03 t = \ln (\frac{2 (15)}{142})$

Step 2.2

Factor $2$ out of $142$ .

$- 0.03 t = \ln (\frac{2 \cdot 15}{2 \cdot 71})$

Step 2.3

Cancel the common factor.

$- 0.03 t = \ln (\frac{2 \cdot 15}{2 \cdot 71})$

Step 2.4

Rewrite the expression.

$- 0.03 t = \ln (\frac{15}{71})$

Step 3

Divide each term in $- 0.03 t = \ln (\frac{15}{71})$ by $- 0.03$ and simplify.

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Step 3.1

Divide each term in $- 0.03 t = \ln (\frac{15}{71})$ by $- 0.03$ .

$\frac{- 0.03 t}{- 0.03} = \frac{\ln (\frac{15}{71})}{- 0.03}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $- 0.03$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{- 0.03 t}{- 0.03} = \frac{\ln (\frac{15}{71})}{- 0.03}$

Step 3.2.1.2

Divide $t$ by $1$ .

$t = \frac{\ln (\frac{15}{71})}{- 0.03}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Move the negative in front of the fraction.

$t = - \frac{\ln (\frac{15}{71})}{0.03}$

Step 3.3.2

Replace $e$ with an approximation.

$t = - \frac{\log_{2.71828182} (\frac{15}{71})}{0.03}$

Step 3.3.3

Divide $15$ by $71$ .

$t = - \frac{\log_{2.71828182} (0.2112676)}{0.03}$

Step 3.3.4

Log base $2.71828182$ of $0.2112676$ is approximately $- 1.55462967$ .

$t = - \frac{- 1.55462967}{0.03}$

Step 3.3.5

Divide $- 1.55462967$ by $0.03$ .

$t = - - 51.82098919$

Step 3.3.6

Multiply $- 1$ by $- 51.82098919$ .

$t = 51.82098919$