Chapter 3 - Review Exercises - Page 514: 86

a) The value of $ k $ is $0.036$. b) The Hispanic resident population in the year 2015 was $60.\text{6 million}$. c) $ t=19$ corresponds to the year 2019.

(a) Consider the function $ A=35.3{{e}^{kt}}$ In $2010$, the population was 50.5 million. For the year $2000$, $ t=0$. The corresponding value for year $2010$ is $ t=10$. Now, compute the value of $ k $. Substitute $ t=10,\ A=50.5$ So, $\begin{align} & A=35.3{{e}^{kt}} \\ & 50.5=35.3{{e}^{k\left( 10 \right)}} \\ & {{e}^{10k}}=\frac{50.5}{35.3} \end{align}$ Take the natural logarithm ${{\log }_{e}}$ on both sides $\begin{align} & {{e}^{10k}}=\frac{50.5}{35.3} \\ & \ln \left( {{e}^{10k}} \right)=\ln \left( \frac{50.5}{35.3} \right) \\ & 10k=\ln \left( \frac{50.5}{35.3} \right) \\ & k=\frac{\ln \left( \frac{50.5}{35.3} \right)}{10} \end{align}$ Now, use the calculator Therefore, the value of $ k $ up to three decimal places is $0.036$. (b) Consider the function as follows: In the year $2010$, the population was 50.5 million. For the year $2000$, $ t=0$. So, for the year $2015$, $ t=15$. Now, calculate the value of population in the year 2015. Put $ t=15,\ k=0.036$ So, $\begin{align} & A=35.3{{e}^{kt}} \\ & A=35.3{{e}^{\left( 0.036 \right)\left( 15 \right)}} \\ & =35.3{{e}^{0.54}} \end{align}$ So, the value is $\begin{align} & A=35.3{{e}^{0.54}} \\ & \approx 60.\text{6 million} \end{align}$ Therefore, the population in the year 2015 is $60.\text{6 million}$. (c) Consider the function as follows: $ A=35.3{{e}^{kt}}$ In the year $2010$, the population was 50.5 million. For the year 2000, $ t=0$. Now, calculate the year when the population reaches 70 million. Put $ A=70,\ k=0.036$ So, $\begin{align} & A=35.3{{e}^{kt}} \\ & 70=35.3{{e}^{\left( 0.036 \right)\left( t \right)}} \\ & {{e}^{0.036t}}=\frac{70}{35.3} \end{align}$ Take the natural logarithm ${{\log }_{e}}$ on both sides as follows: $\begin{align} & {{e}^{0.036t}}=\frac{70}{35.3} \\ & \ln \left( {{e}^{0.036t}} \right)=\ln \left( \frac{70}{35.3} \right) \\ & 0.036t=\ln \left( \frac{70}{35.3} \right) \\ & t=\frac{\ln \left( \frac{70}{35.3} \right)}{0.036} \end{align}$ Now, use the calculator to obtain the value of $t$. So, the value is: $\begin{align} & t=\frac{\ln \left( \frac{70}{35.3} \right)}{0.036} \\ & =19.017 \\ & \approx 19 \end{align}$ Therefore, $t=19$ corresponds to the year 2019.