Question

A population, P(t) (in millions) in year t, increases exponentially. Suppose P(9)=16 and P(18)=24. a) Find a formula for the population in the form P(t)=abt. Enter the values you found for a and b in your formula in the blanks below. Round your values to 4 decimal places.

Answer 1

Answer:$\frac{32}{3}\left ( 1.0460\right )^t$

Step-by-step explanation:

Given

Population changes exponentially

$P\left ( t\right )=ab^t$

$P\left ( 9\right )=16$------1

$P\left ( 9\right )=16=ab^8$

$P\left ( 18\right =24=ab^{18}$-----2

divide 1 & 2 we get

$\frac{24}{16}=\frac{b^{18}}{b^9}$

$\frac{3}{2}=b^9$

Substitute $b^9$ in 1 we get

$a=\frac{32}{3}$

Thus $P\left ( t\right )=\frac{32}{3}\left ( 1.0460\right )^t$

Answer 2

Answer:

$P(t)=10.6667(1.0461)^t$

$a\approx 10.6667$

$b\approx 1.0461$

Step-by-step explanation:

It is given that a population, P(t) (in millions) in year t, increases exponentially.

The given values of he function are P(9)=16 and P(18)=24.

The formula for the population in the form

$P(t)=ab^t$

where, a is the initial population and b is growth factor.

P(9)=16 means P(t)=16 at x=9.

$16=ab^9$          .... (1)

P(18)=24 means P(t)=24 at x=18.

$24=ab^{18}$          .... (2)

Divide equation (2) by equation (1).

$\dfrac{24}{16}=\dfrac{ab^{18}}{ab^9}$

$1.5=b^9$

$(1.5)^{\frac{1}{9}}=b$

$b\approx 1.0461$

Substitute $b^9=1.5$ in equation (1).

$16=a(1.5)$

Divide both sides by 1.5.

$\dfrac{16}{1.5}=a$

$a\approx 10.6667$

Substitute the values of a and b in the given formula.

$P(t)=10.6667(1.0461)^t$

Therefore, the formula for population is $P(t)=10.6667(1.0461)^t$.