Calculus Examples

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

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

Step 1

Rewrite the equation as

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

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

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

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

Step 2

Reduce the expression

\frac{30}{142}

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

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Step 2.1

Factor

2

$2$ out of

30

$30$ .

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

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

Step 2.2

Factor

2

$2$ out of

142

$142$ .

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

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

Step 2.3

Cancel the common factor.

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

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

Step 2.4

Rewrite the expression.

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

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

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

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

Step 3

Divide each term in

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

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

- 0.03

$- 0.03$ and simplify.

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Step 3.1

Divide each term in

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

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

- 0.03

$- 0.03$ .

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

$- 0.03$ .

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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}

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

Step 3.2.1.2

Divide

t

$t$ by

1

$1$ .

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

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

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

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

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

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

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}

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

Step 3.3.2

Replace

e

$e$ with an approximation.

t = - \frac{{log}_{2.71828182} (\frac{15}{71})}{0.03}

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

Step 3.3.3

Divide

15

$15$ by

71

$71$ .

t = - \frac{{log}_{2.71828182} (0.2112676)}{0.03}

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

Step 3.3.4

Log base

2.71828182

$2.71828182$ of

0.2112676

$0.2112676$ is approximately

- 1.55462967

$- 1.55462967$ .

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

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

Step 3.3.5

Divide

- 1.55462967

$- 1.55462967$ by

0.03

$0.03$ .

t = - - 51.82098919

$t = - - 51.82098919$

Step 3.3.6

Multiply

- 1

$- 1$ by

- 51.82098919

$- 51.82098919$ .

t = 51.82098919

$t = 51.82098919$

t = 51.82098919

$t = 51.82098919$

t = 51.82098919

$t = 51.82098919$