Question

In 2000 the population of a country reached 1 ​billion, and in 2025 it is projected to be 1.2 billion. ​(a) Find values for C and a so that ​P(x)equalsCa Superscript x minus 2000 models the population of a country in year x. ​(b) Estimate the​ country's population in 2010. ​(c) Use P to determine the year when the​ country's population might reach 1.4 billion. ​(a) Cequals nothing ​(Type an integer or decimal rounded to five decimal places as​ needed.)

Answer 1

Answer:

(a) The value of C is 1.

(b) In 2010, the population would be 1.07555 billions.

(c) In 2047, the population would be 1.4 billions.

Step-by-step explanation:

(a) Here, the given function that shows the population(in billions) of the country in year x,

$P(x)=Ca^{x-2000}$

So, the population in 2000,

$P(2000)=Ca^{2000-2000}$

$=Ca^{0}$

$=C$

According to the question,

$P(2000)=1$

$\implies C=1$

(b) Similarly,

The population in 2025,

$P(2025)=Ca^{2025-2000}$

$=Ca^{25}$

$=a^{25}$                    (∵ C = 1)

Again according to the question,

$P(2025)=1.2$

$a^{25}=1.2$

Taking ln both sides,

$\ln a^{25}=\ln 1.2$

$25\ln a = \ln 1.2$

$\ln a = \frac{\ln 1.2}{25}\approx 0.00729$

$a=e^{0.00729}=1.00731$

Thus, the function that shows the population in year x,

$P(x)=(1.00731)^{x-2000}$     ...... (1)

The population in 2010,

$P(2010)=(1.00731)^{2010-2000}=(1.00731)^{10}=1.07555$          

Hence, the population in 2010 would be 1.07555 billions.

(c) If population P(x) = 1.4 billion,

Then, from equation (1),

$1.4=(1.00731)^{x-2000}$

$\ln 1.4=(x-2000)\ln 1.00731$

$0.33647 = (x-2000)0.00728$

$0.33647 = 0.00728x-14.56682$

$0.33647 + 14.56682 = 0.00728x$

$14.90329 = 0.00728x$

$\implies x=\frac{14.90329}{0.00728}\approx 2047$

Therefore, the country's population might reach 1.4 billion in 2047.