Respuesta :
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,
[tex]P(x)=Ca^{x-2000}[/tex]
So, the population in 2000,
[tex]P(2000)=Ca^{2000-2000}[/tex]
[tex]=Ca^{0}[/tex]
[tex]=C[/tex]
According to the question,
[tex]P(2000)=1[/tex]
[tex]\implies C=1[/tex]
(b) Similarly,
The population in 2025,
[tex]P(2025)=Ca^{2025-2000}[/tex]
[tex]=Ca^{25}[/tex]
[tex]=a^{25}[/tex] (∵ C = 1)
Again according to the question,
[tex]P(2025)=1.2[/tex]
[tex]a^{25}=1.2[/tex]
Taking ln both sides,
[tex]\ln a^{25}=\ln 1.2[/tex]
[tex]25\ln a = \ln 1.2[/tex]
[tex]\ln a = \frac{\ln 1.2}{25}\approx 0.00729[/tex]
[tex]a=e^{0.00729}=1.00731[/tex]
Thus, the function that shows the population in year x,
[tex]P(x)=(1.00731)^{x-2000}[/tex] ...... (1)
The population in 2010,
[tex]P(2010)=(1.00731)^{2010-2000}=(1.00731)^{10}=1.07555[/tex]
Hence, the population in 2010 would be 1.07555 billions.
(c) If population P(x) = 1.4 billion,
Then, from equation (1),
[tex]1.4=(1.00731)^{x-2000}[/tex]
[tex]\ln 1.4=(x-2000)\ln 1.00731[/tex]
[tex]0.33647 = (x-2000)0.00728[/tex]
[tex]0.33647 = 0.00728x-14.56682[/tex]
[tex]0.33647 + 14.56682 = 0.00728x[/tex]
[tex]14.90329 = 0.00728x[/tex]
[tex]\implies x=\frac{14.90329}{0.00728}\approx 2047[/tex]
Therefore, the country's population might reach 1.4 billion in 2047.
