Chapter 1.7, Problem 30E

To determine

The graph of the given exponential functions

The value of x where it is equal to 7.0

The doubling time and the half time

The value of x for which the function is 3.5

The value of x for which the function is 14.0

Answer to Problem 30E

  $\begin{array}{l}x=-21.242\\ t_h=3.46\\ x=-17.77\\ x=-24.71\end{array}$

Explanation of Solution

Given:

The given function is

  $0.1e^{-0.2x}$

Calculation:

The graph of the function will be

  

The value of x for which the function is 7,

  $\begin{array}{l}0.1e^{-0.2x}=7\\ \ln \left(0.1e^{-0.2x}\right)=\ln \left(7\right)\\ \ln \left(0.1\right)+\ln \left(e^{-0.2x}\right)=\ln \left(7\right)\\ \ln \left(e^{-0.2x}\right)=\ln \left(7\right)-\ln \left(0.1\right)\\ \ln e=1\\ \ln \left(7/0.1\right)=4.248\end{array}$

Therefore,

  $\begin{array}{l}\ln \left(e^{-0.2x}\right)=\ln \left(7\right)-\ln \left(0.1\right)\\ -0.2x=\ln \left(7/0.1\right)\\ -0.2x=4.248\\ x=-21.242\end{array}$

The function is decreasing; the half time can be calculated. The general for half time is

  $\begin{array}{l}{t}_h=\frac{\ln \left(0.5\right)}{\alpha }\\ t_h=-\frac{0.693}{-0.2}\\ t_h=3.46\end{array}$

The value of x for which the function is 3.5,

  $\begin{array}{l}0.1e^{-0.2x}=3.5\\ \ln \left(0.1e^{-0.2x}\right)=\ln \left(3.5\right)\\ \ln \left(0.1\right)+\ln \left(e^{-0.2x}\right)=\ln \left(3.5\right)\\ \ln \left(e^{-0.2x}\right)=\ln \left(3.5\right)-\ln \left(0.1\right)\\ \ln e=1\\ \ln \left(3.5/0.1\right)=0.356\end{array}$

Therefore,

  $\begin{array}{l}\ln \left(e^{-0.2x}\right)=\ln \left(3.5\right)-\ln \left(0.1\right)\\ -0.2x=\ln \left(3.5/0.1\right)\\ -0.2x=0.356\\ x=-17.77\end{array}$

The value of x for which the function is 14,

  $\begin{array}{l}0.1e^{-0.2x}=14\\ \ln \left(0.1e^{-0.2x}\right)=\ln \left(14\right)\\ \ln \left(0.1\right)+\ln \left(e^{-0.2x}\right)=\ln \left(14\right)\\ \ln \left(e^{-0.2x}\right)=\ln \left(14\right)-\ln \left(0.1\right)\\ \ln e=1\end{array}$

Therefore,

  $\begin{array}{l}\ln \left(e^{-0.2x}\right)=\ln \left(14\right)-\ln \left(0.1\right)\\ -0.2x=\ln \left(14/0.1\right)\\ -0.2x=4.942\\ x=-24.71\end{array}$