Chapter 7.4, Problem 5E

a.

To determine

To predict : the world population in $1900$ and $1950$ by using the exponential model of $1750$ and $1800$

Answer to Problem 5E

The $y\left(150\right)=1508.08$ & $y\left(200\right)=1870.78$

Explanation of Solution

Given information : The exponential model is

Year	Population
1750	790
1800	980

Calculation :

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(50\right)=790\times e^{k\times 50}\\ \frac{980}{790}=e^{k\times 50}\\ \ln \left(\frac{98}{79}\right)=k\times 50\\ \frac{1}{50}\times \ln \left(\frac{98}{79}\right)=k\end{array}$

Let $y\left(0\right)$

  $=790$ or the population in $1750$

Let $y\left(50\right)=$

  $980$ or the population of $1800$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=790\times e^{\frac{1}{50}\times \ln \left(\frac{98}{79}\right)\times t}\\ y\left(t\right)=790\times e^{\frac{t}{50}\times \ln \left(\frac{98}{79}\right)}\\ y\left(t\right)=790\times {\left(\frac{98}{79}\right)}^{\frac{t}{50}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(150\right)=790\times {\left(\frac{98}{79}\right)}^{\frac{150}{50}}\\ y\left(150\right)=1508.08\\ y\left(200\right)=790\times {\left(\frac{98}{79}\right)}^{\frac{200}{50}}\\ y\left(200\right)=1870.78\\ \text{At }1900,t=150\text{ because }1900-1750=150\\ \text{At }1950,t=200\text{ because }1950-1750=200\\ y\left(150\right)=1508.08\text{ v/s actual population }1650\\ y\left(200\right)=1870.78\text{ v/s actual population }2560\end{array}$

b.

To determine

To predict : the world population in $1950$ by using the exponential model of $1850$ and $1900$

Answer to Problem 5E

The $y\left(100\right)=2160.71$

Explanation of Solution

Given information : The exponential model is

Year	Population
1850	1260
1900	1650

Calculation :

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(50\right)=1260\times e^{k\times 50}\\ \frac{1650}{1260}=e^{k\times 50}\\ \ln \left(\frac{165}{126}\right)=k\times 50\\ \frac{1}{50}\times \ln \left(\frac{55}{42}\right)=k\end{array}$

Let $y\left(0\right)$

  $=1260$ or the population in $1850$

Let $y\left(50\right)=$

  $1650$ or the population of $1900$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=1260\times e^{\frac{1}{50}\times \ln \left(\frac{55}{42}\right)\times t}\\ y\left(t\right)=1260\times e^{\frac{t}{50}\times \ln \left(\frac{55}{42}\right)}\\ y\left(t\right)=1260\times {\left(\frac{55}{42}\right)}^{\frac{t}{50}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(100\right)=1260\times {\left(\frac{55}{42}\right)}^{\frac{100}{50}}\\ y\left(100\right)=2160.71\end{array}$

  $\begin{array}{l}\text{At }1950,t=100\text{ because }1950-1850=100\\ y\left(100\right)=2160.71\text{ v/s actual population }2560\end{array}$

c.

To determine

To predict : the world population in $2000$ by using the exponential model of $1900$ and $1950$

Answer to Problem 5E

The $y\left(100\right)=3971.88$

Explanation of Solution

Given information : The exponential model is

Year	Population
1900	1650
1950	2560

Calculation :

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(50\right)=1650\times e^{k\times 50}\\ \frac{2560}{1650}=e^{k\times 50}\\ \ln \left(\frac{256}{165}\right)=k\times 50\\ \frac{1}{50}\times \ln \left(\frac{256}{165}\right)=k\end{array}$

Let $y\left(0\right)$

  $=1650$ or the population in $1900$

Let $y\left(50\right)=$

  $2560$ or the population of $1950$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=1650\times e^{\frac{1}{50}\times \ln \left(\frac{256}{165}\right)\times t}\\ y\left(t\right)=1650\times e^{\frac{t}{50}\times \ln \left(\frac{256}{165}\right)}\\ y\left(t\right)=1650\times {\left(\frac{256}{165}\right)}^{\frac{t}{50}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(100\right)=1650\times {\left(\frac{256}{165}\right)}^{\frac{100}{50}}\\ y\left(100\right)=3971.88\end{array}$

  $\begin{array}{l}\text{At }2000,t=100\text{ because }2000-1900=100\\ y\left(100\right)=3971.88\text{ v/s actual population }6080\end{array}$