Chapter 0.3, Problem 3QQ

To determine

The best approximation time for $1$ gram remaining of the doze.

Answer to Problem 3QQ

The best approximation time is option $\left(d\right)19$ .

Explanation of Solution

Given:

Half life of the substance $=8\text{hr}$

Initially the weight of the substance is $=5\text{gram}$

Find the best approximation time for $1$ gram remaining of the doze.

Calculation:

Given that,

Half-life of the substance $=8\text{hr}$

Initially the weight of the substance is $=5\text{gram}$

Apply the half-life formula:

  $N\left(T\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{\left(\frac{1}{2}\right)}}}$

Here, $N\left(t\right)=$ Quantity of the substance remaining

  $N_0=$ Initial quantity of the substance

  $t=$ Time elapsed

  $t_{\left(\frac{1}{2}\right)}=$ Half-life of the substance

Now, substitute half life $t_{\left(\frac{1}{2}\right)}=8$ and initial quantity $N_0=5$ in the formula and quantity of the remaining substance value $N\left(t\right)=1$ into the formula $N\left(T\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{\left(\frac{1}{2}\right)}}}$ :

  $1=5{\left(\frac{1}{2}\right)}^{\frac{t}{8}}$

Rewrite the equation:

  $5{\left(\frac{1}{2}\right)}^{\frac{t}{8}}=1$

Divide both sides by $5$ and simplify:

  ${\left(\frac{1}{2}\right)}^{\frac{t}{8}}=\frac{1}{5}$

Rewrite it again:

  $\frac{1^{\left(\frac{t}{8}\right)}}{2^{\left(\frac{t}{8}\right)}}=\frac{1}{5}$

One to any power is one:

  $\frac{1}{2^{\left(\frac{t}{8}\right)}}=\frac{1}{5}$

Cross multiply $2^{\left(\frac{t}{8}\right)}$ with $1$ :

  $\begin{array}{l}1=\frac{2^{\frac{t}{8}}}{5}\\ 5=2^{\frac{t}{8}}\end{array}$

Take the natural logarithm of both sides of the equation to remove the variable from

the exponent:

  $\ln \left(2^{\frac{t}{8}}\right)=\ln \left(5\right)$

Expand $\ln \left(2^{\frac{t}{8}}\right)$ by moving $\frac{t}{8}$ outside the logarithm:

  $\frac{t}{8}\ln \left(2\right)=\ln \left(5\right)$

Combine $\frac{t}{8}$ and $\ln \left(2\right)$ :

  $\frac{t\ln \left(2\right)}{8}=\ln \left(5\right)$

Solve for $t$ :

  $t=\frac{8\ln \left(5\right)}{\ln \left(2\right)}$

In decimal form the value of $t$ is:

  $\begin{array}{l}t=18.57542475\\ t\approx 19\end{array}$

Thus, the best approximation time for $1$ gram remaining of the doze is option $\left(d\right)19$ .