Chapter 5, Problem 21CT

To determine

The days that will be taken for there to be $2$ mg remaining, where a $50$ -mg sample of a radioactive substance decays to $34$ mg after $30$ days.

Answer to Problem 21CT

Solution:

  $250.39$ days will be taken for there to be $2$ mg remaining.

Explanation of Solution

Given information:

A $50$ -mg sample of a radioactive substance decays to $34$ mg after $30$ days.

The amount $A$ of radioactive substance at time $t$ is $A\left(t\right)=A_0{e}^{kt}$ , where $A_0$ is the original amount of radioactive substance, and $k$ is a negative number.

Here, $A_0=50$ mg is the original amount of radioactive substance.

It is given that at $t=30$ , $A\left(30\right)=34$ mg.

Substitute $A_0=50$ , $t=30$ , and $A\left(30\right)=34$ ,

  $\Rightarrow 34=50e^{k\left(30\right)}$

Dividing both sides by $40$ ,

  $\Rightarrow \frac{34}{50}=e^{k\left(30\right)}$

By using $x=e^y$ if and only if $y=\ln x$ ,

  $\Rightarrow 30k=\ln \left(\frac{34}{50}\right)$

  $\Rightarrow 30k=\ln \left(\frac{17}{25}\right)$

  $\Rightarrow k=\frac{\ln \left(\frac{17}{25}\right)}{30}$

  $\Rightarrow k=-0.01285541603$

Exponential model becomes $A\left(t\right)=50e^{-0.01285541603t}$ .

Now find $t$ when $A\left(t\right)=2$ ,

  $\Rightarrow 2=50e^{-0.01285541603t}$

Dividing both sides by $50$ ,

  $\Rightarrow \frac{2}{50}=e^{-0.01285541603t}$

  $\Rightarrow \frac{1}{25}=e^{-0.01285541603t}$

By using $x=e^y$ if and only if $y=\ln x$ ,

  $\Rightarrow -0.01285541603t=\ln \left(\frac{1}{25}\right)$

  $\Rightarrow -0.01285541603t=-3.218875825$

Dividing both sides by $-0.01285541603$ ,

  $\Rightarrow t=\frac{-3.218875825}{-0.01285541603}$

  $\Rightarrow t=250.3906383$

  $\Rightarrow t\approx 250.39$

Therefore, the $250.39$ days will be taken for there to be $2$ mg remaining.