Chapter 3.2, Problem 34E

Solution

(a).

To express: The population the amount of substance remaining as a function of time $t$ .

The population the amount of substance remaining as a function of time $t$ is $m\left(t\right)=3.5\times {\left(\frac{1}{2}\right)}^{\frac{t}{65}}$

Given information:

The half-life of a certain radioactive substance is $65$ days. There are $3.5g$ present initially.

Calculation:

It is initial mass of $3.5g$ halves every $65$ days, then every $65$ days it is multiplied by $\frac{1}{2}$ :

  $\underset{}{{\displaystyle m=3.5\cdot \underset{\text{once every 65 days}}{\underbrace{\frac{1}{2}\cdot \frac{1}{2}\cdot \cdot \cdot \frac{1}{2}}}}}$

If $t$ days have passed, then the mass will have halved $\frac{t}{65}$ times. Thus, the function is:

  $m\left(t\right)=3.5\times {\left(\frac{1}{2}\right)}^{\frac{t}{65}}$

(b).

To find: When will there be less than $1g$ remaining?

The required time is approximately $t=117.5$ days.

Given information:

The half-life of a certain radioactive substance is $65$ days. There are $3.5g$ present initially.

Calculation:

The function is $m\left(t\right)=3.5\times {\left(\frac{1}{2}\right)}^{\frac{t}{65}}$

To simplify notation, write the function as:

  $m\left(t\right)=3.5\times 2^{-\frac{t}{65}}$

If there is $1$ gram of mass left, then $m\left(t\right)=1$ .

  $3.5\times 2^{-\frac{t}{65}}=1$

Taking natural log to the both sides:

  $\begin{array}{l}\ln \left[3.5\times 2^{-\frac{t}{65}}\right]=\ln 1\\ \ln 3.5-\frac{t}{65}\ln 2=0\\ t=\frac{65\ln 3.5}{\ln 2}\\ t\approx 117.478070\\ t\approx 117.5\text{ days}\end{array}$

Hence, it must round up to $t=117.5$ days.