Answer :
We are given that the population in 1985 was [tex]$145$[/tex] million and in 1995, [tex]$10$[/tex] years later, it was [tex]$190$[/tex] million. We will use the exponential growth model
[tex]$$
P(t) = P_0 e^{rt},
$$[/tex]
where:
- [tex]$P(t)$[/tex] is the population after time [tex]$t$[/tex],
- [tex]$P_0$[/tex] is the initial population,
- [tex]$r$[/tex] is the annual growth rate,
- [tex]$t$[/tex] is the time (in years).
Step 1. Find the annual growth rate, [tex]$r$[/tex].
Using the data from 1985 to 1995, let the time elapsed be [tex]$t = 10$[/tex] years. Then we have
[tex]$$
190 = 145 \, e^{10r}.
$$[/tex]
To solve for [tex]$r$[/tex], divide both sides by [tex]$145$[/tex]:
[tex]$$
e^{10r} = \frac{190}{145}.
$$[/tex]
Now, take the natural logarithm of both sides:
[tex]$$
10r = \ln\left(\frac{190}{145}\right).
$$[/tex]
Thus, the annual growth rate is
[tex]$$
r = \frac{1}{10}\ln\left(\frac{190}{145}\right).
$$[/tex]
When computed, this value is approximately [tex]$0.027$[/tex].
Step 2. Express the population in 2005.
The year 2005 is [tex]$10$[/tex] years after 1995. Since the population in 1995 is given as [tex]$190$[/tex] million, we will use the exponential growth formula starting from 1995:
[tex]$$
P_{\text{2005}} = 190 \, e^{r \cdot 10}.
$$[/tex]
Substitute the approximate value [tex]$r \approx 0.027$[/tex]:
[tex]$$
P_{\text{2005}} = 190 \, e^{0.027 \times 10}.
$$[/tex]
This is the expression that represents the population in 2005.
Step 3. Conclude the correct choice.
Comparing with the options provided:
A. [tex]$P=145\,e^{(0.027)(10)}$[/tex]
B. [tex]$P=145\,e^{(0.027)(20)}$[/tex]
C. [tex]$P=190\,e^{(0.027)(10)}$[/tex]
D. [tex]$P=190\,e^{(0.027)(20)}$[/tex]
We see that the expression that correctly represents the population in 2005 is
[tex]$$
P = 190 \, e^{(0.027)(10)},
$$[/tex]
which corresponds to Option C.
[tex]$$
P(t) = P_0 e^{rt},
$$[/tex]
where:
- [tex]$P(t)$[/tex] is the population after time [tex]$t$[/tex],
- [tex]$P_0$[/tex] is the initial population,
- [tex]$r$[/tex] is the annual growth rate,
- [tex]$t$[/tex] is the time (in years).
Step 1. Find the annual growth rate, [tex]$r$[/tex].
Using the data from 1985 to 1995, let the time elapsed be [tex]$t = 10$[/tex] years. Then we have
[tex]$$
190 = 145 \, e^{10r}.
$$[/tex]
To solve for [tex]$r$[/tex], divide both sides by [tex]$145$[/tex]:
[tex]$$
e^{10r} = \frac{190}{145}.
$$[/tex]
Now, take the natural logarithm of both sides:
[tex]$$
10r = \ln\left(\frac{190}{145}\right).
$$[/tex]
Thus, the annual growth rate is
[tex]$$
r = \frac{1}{10}\ln\left(\frac{190}{145}\right).
$$[/tex]
When computed, this value is approximately [tex]$0.027$[/tex].
Step 2. Express the population in 2005.
The year 2005 is [tex]$10$[/tex] years after 1995. Since the population in 1995 is given as [tex]$190$[/tex] million, we will use the exponential growth formula starting from 1995:
[tex]$$
P_{\text{2005}} = 190 \, e^{r \cdot 10}.
$$[/tex]
Substitute the approximate value [tex]$r \approx 0.027$[/tex]:
[tex]$$
P_{\text{2005}} = 190 \, e^{0.027 \times 10}.
$$[/tex]
This is the expression that represents the population in 2005.
Step 3. Conclude the correct choice.
Comparing with the options provided:
A. [tex]$P=145\,e^{(0.027)(10)}$[/tex]
B. [tex]$P=145\,e^{(0.027)(20)}$[/tex]
C. [tex]$P=190\,e^{(0.027)(10)}$[/tex]
D. [tex]$P=190\,e^{(0.027)(20)}$[/tex]
We see that the expression that correctly represents the population in 2005 is
[tex]$$
P = 190 \, e^{(0.027)(10)},
$$[/tex]
which corresponds to Option C.