Chapter 6.6, Problem 18E

To determine

To calculate: .The number of years when only $5$ grams of radium will be present in a container.

Answer to Problem 18E

The required number of years are $6976.74$ .

Explanation of Solution

Given Information:

The given function is $R=100e^{-0.00034t}$ .

Where $R$ is the amount of radium (in grams) and $t$ is year.

Formula Used:

Power rule of logarithmic functions:

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variables and $p$ is the power of variables.

Calculation:

Consider the function $R=100e^{-0.00034t}$ .

Substitute $5$ for $R$ in the above equation.

  $5=100e^{-0.00043t}$

Simplify the above equation.

  $5=100e^{-0.00043t}$

Divide each side by $100$ in the above equation.

  $\begin{array}{c}\frac{5}{100}=\frac{100e^{-0.00043t}}{100}\\ 0.05=e^{-0.00043t}\end{array}$

Take natural log each side with base $e$ of the above equation.

  ${\ln }_e\left(0.05\right)={\ln }_e{e}^{-0.00043t}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  ${\ln }_e\left(0.05\right)=\left(-0.00043t\right){\ln }_{e}e$

Substitute $1$ for ${\ln }_{e}e$ in the above equations.

  $\begin{array}{c}{\ln }_e\left(0.05\right)=\left(-0.00043t\right)\left(1\right)\\ {\ln }_e\left(0.05\right)=\left(-0.00043t\right)\end{array}$

Substitute $-3$ for ${\ln }_e\left(0.05\right)$ in the above equation.

  $-3=\left(-0.00043t\right)$

Simplify the above equation.

  $\begin{array}{c}-3=\left(-0.00043\right)t\\ t=\frac{-3.00}{\left(-0.00043\right)}\\ t=6979.74\end{array}$

Hence, the required number of years are $6976.74$