Chapter 1.5, Problem 46E

(a)

To determine

To express: The amount of substance remaining as a function of time.

Answer to Problem 46E

The amount of substance remaining as a function of time is $f\left(t\right)=8{\left(\frac{1}{2}\right)}^{\frac{t}{12}}$.

Explanation of Solution

Given information: The half-life of substance is $12$ hours and the initial amount of substance is $8$ grams.

Calculation:

The formula of exponential growth is $y=a{b}^x$.

Here, $b$ will be one-half because half-life is given and $a$ is the initial number.

So, $a=8$ because it is given that initial amount is 8 grams. The $x$ in the exponent is replaced with $t$ because it is specified that $t$ as the variable and t is divided by twelve because the half life is twelve hours.

The amount of substance remaining as a function of time can be expressed as follows:

  $f\left(t\right)=8{\left(\frac{1}{2}\right)}^{\frac{t}{12}}$

Therefore, the amount of substance remaining as a function of time can be expressed as $f\left(t\right)=8{\left(\frac{1}{2}\right)}^{\frac{t}{12}}$.

(b)

To determine

To find: The required time such that substance remains 1 gram.

Answer to Problem 46E

The substance will remain 1 gram after $36$ hours.

Explanation of Solution

Given information: The half-life of substance is $12$ hours and initial amount of substance is $8$ grams.

Calculation:

Consider the function $f\left(t\right)=8{\left(\frac{1}{2}\right)}^{\frac{t}{12}}$.

Substitute 1 for $f\left(t\right)$ in the above equation.

  $\begin{array}{c}1=8{\left(\frac{1}{2}\right)}^{\frac{t}{12}}\\ {\left(\frac{1}{2}\right)}^{\frac{t}{12}}=\frac{1}{8}\\ {\left(\frac{1}{2}\right)}^{\frac{t}{12}}={\left(\frac{1}{2}\right)}^3\\ \frac{t}{12}=3\end{array}$

Simplify further.

  $\begin{array}{c}t=12\left(3\right)\\ =36\text{hours}\end{array}$

Therefore, the substance will remain 1 gram after $36$ hours.