Answer:
(a) The growth factor is 1.9.
(b) The population after 10 and 15 years are 18393 and 455434 respectively.
(c) \(p=30(1.9)^n\)
(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
\(f(x)=ab^x\) .... (1)
where, a is initial value and b is growth factor.
Consider any two point from the given table. (0,30) and (1,57).
\(30=ab^0\)
\(30=a\)
The value of a is 30.
\(57=ab^1\)
\(57=30b\)
Divide both sides by 30.
\(\frac{57}{30}=b\)
\(1.9=b\)
Substitute a=30 and b=1.9 in equation (1).
\(f(x)=30(1.9)^x\) .... (2)
(a)
\(b=1.9\)
Therefore the growth factor is 1.9.
(b)
Substitute x=10 in equation (2), to find the population after 10 years.
\(f(10)=30(1.9)^{10}\)
\(f(10)=18393.1987734\)
\(f(10)\approx 18393\)
Therefore the population after 10 years is 18393.
Substitute x=15 in equation (2), to find the population after 15 years.
\(f(15)=30(1.9)^{15}\)
\(f(15)=455433.810896\)
\(f(15)\approx 455434\)
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
\(p=30(1.9)^n\)
(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.
\(1000000<30(1.9)^t\)
Divide both sides by 30.
\(\frac{100000}{3}<(1.9)^t\)
Taking ln both sides.
\(\ln \frac{100000}{3}<\ln (1.9)^t\)
\(\ln \frac{100000}{3}
Divide both sides by ln(1.9).
\(\frac{\ln \frac{100000}{3}}{\ln (1.9)}
\(16.2253643713
Therefore in 17 years population exceed one million.
