The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.

Let \(\displaystyle{A}_{{0}}\) and r be the initial population & growth rate of population respectively. Then population in California t years after 1990 is,

\(\displaystyle{P}{\left({t}\right)}={A}_{{0}}{e}^{{{r}{t}}}\) (1)

a. Since the population in 1990 is 29.76 million, therefore \(\displaystyle{A}_{{0}}={29.76}\)

In 2000, the population is 33.87 million. That is,

\(\displaystyle{P}{\left({t}\right)}={33.87}\) when \(\displaystyle{t}={10}\)

Therefore from equation (1), we get

\(\displaystyle{33.87}={29.76}{e}^{{{10}{r}}}\)

\(\displaystyle\Rightarrow{e}^{{{10}{r}}}={1.13810483}\)

\(\displaystyle\Rightarrow{10}{r}={\ln{{\left({1.13810483}\right)}}}\)

\(\displaystyle\Rightarrow{r}={\frac{{{\ln{{\left({1.13810483}\right)}}}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{r}={0.012936}\)

Therefore the population t years after 1990 is,

\(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\)

b) The double of 29.76 million is 59.52 million.

So put P(t)=59.52 in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\), we get

\(\displaystyle{59.52}={29.76}{e}^{{{0.012936}{t}}}\)

\(\displaystyle\Rightarrow{e}^{{{0.012936}{t}}}={2}\)

\(\displaystyle\Rightarrow{0.012936}{t}={\ln{{\left({2}\right)}}}\)

\(\displaystyle\Rightarrow{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.012936}}}}\)

\(\displaystyle\Rightarrow{t}={53.5828}\)

Hence the population will double in 53.5828 years.

c)Since 2010−1990=20, therefore to predict the population in 2010, put t=20 in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\) , we get

\(\displaystyle{P}{\left({20}\right)}={29.76}{e}^{{{0.012936}{\left({20}\right)}}}\)

\(\displaystyle={38.5472}\) million

The population in California in 2010 is 37.3 million, therefore the actual population is approximately 1 million less than the obtained population.

Let \(\displaystyle{A}_{{0}}\) and r be the initial population & growth rate of population respectively. Then population in California t years after 1990 is,

\(\displaystyle{P}{\left({t}\right)}={A}_{{0}}{e}^{{{r}{t}}}\) (1)

a. Since the population in 1990 is 29.76 million, therefore \(\displaystyle{A}_{{0}}={29.76}\)

In 2000, the population is 33.87 million. That is,

\(\displaystyle{P}{\left({t}\right)}={33.87}\) when \(\displaystyle{t}={10}\)

Therefore from equation (1), we get

\(\displaystyle{33.87}={29.76}{e}^{{{10}{r}}}\)

\(\displaystyle\Rightarrow{e}^{{{10}{r}}}={1.13810483}\)

\(\displaystyle\Rightarrow{10}{r}={\ln{{\left({1.13810483}\right)}}}\)

\(\displaystyle\Rightarrow{r}={\frac{{{\ln{{\left({1.13810483}\right)}}}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{r}={0.012936}\)

Therefore the population t years after 1990 is,

\(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\)

b) The double of 29.76 million is 59.52 million.

So put P(t)=59.52 in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\), we get

\(\displaystyle{59.52}={29.76}{e}^{{{0.012936}{t}}}\)

\(\displaystyle\Rightarrow{e}^{{{0.012936}{t}}}={2}\)

\(\displaystyle\Rightarrow{0.012936}{t}={\ln{{\left({2}\right)}}}\)

\(\displaystyle\Rightarrow{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.012936}}}}\)

\(\displaystyle\Rightarrow{t}={53.5828}\)

Hence the population will double in 53.5828 years.

c)Since 2010−1990=20, therefore to predict the population in 2010, put t=20 in \(\displaystyle{P}{\left({t}\right)}={29.76}{e}^{{{0.012936}{t}}}\) , we get

\(\displaystyle{P}{\left({20}\right)}={29.76}{e}^{{{0.012936}{\left({20}\right)}}}\)

\(\displaystyle={38.5472}\) million

The population in California in 2010 is 37.3 million, therefore the actual population is approximately 1 million less than the obtained population.