Answer:
(a) The growth factor is 1.9.
(b) The population after 10 and 15 years are 18393 and 455434 respectively.
(c) [tex]p=30(1.9)^n[/tex]
(d) In 17 years population exceed one million.
Step-by-step explanation:
Given table represents an exponential function.
The general form of an exponential function is
[tex]f(x)=ab^x[/tex] .... (1)
where, a is initial value and b is growth factor.
Consider any two point from the given table. (0,30) and (1,57).
[tex]30=ab^0[/tex]
[tex]30=a[/tex]
The value of a is 30.
[tex]57=ab^1[/tex]
[tex]57=30b[/tex]
Divide both sides by 30.
[tex]\frac{57}{30}=b[/tex]
[tex]1.9=b[/tex]
Substitute a=30 and b=1.9 in equation (1).
[tex]f(x)=30(1.9)^x[/tex] .... (2)
(a)
[tex]b=1.9[/tex]
Therefore the growth factor is 1.9.
(b)
Substitute x=10 in equation (2), to find the population after 10 years.
[tex]f(10)=30(1.9)^{10}[/tex]
[tex]f(10)=18393.1987734[/tex]
[tex]f(10)\approx 18393[/tex]
Therefore the population after 10 years is 18393.
Substitute x=15 in equation (2), to find the population after 15 years.
[tex]f(15)=30(1.9)^{15}[/tex]
[tex]f(15)=455433.810896[/tex]
[tex]f(15)\approx 455434[/tex]
Therefore the population after 15 years is 455434.
(c)
The equation of elk population p for any year n after the elk were first counted is
[tex]p=30(1.9)^n[/tex]
(d)
We need to find the number of years after that the population exceed one million.
Let in t years the population exceed one million.
[tex]1000000<30(1.9)^t[/tex]
Divide both sides by 30.
[tex]\frac{100000}{3}<(1.9)^t[/tex]
Taking ln both sides.
[tex]\ln \frac{100000}{3}<\ln (1.9)^t[/tex]
[tex]\ln \frac{100000}{3}<t\ln (1.9)[/tex]
Divide both sides by ln(1.9).
[tex]\frac{\ln \frac{100000}{3}}{\ln (1.9)}<t[/tex]
[tex]16.2253643713<t[/tex]
Therefore in 17 years population exceed one million.
