Question

In Youngtown, the population reached 4420 people in 1993. In 1998, there were 5600 residents. What was the population in the year 2000?

Answer 1

Population in `2000` was `6156`

Explanation:

Assuming population growth is an exponential function.

Initial population in 1993 is `P_0=4420`

The growth function is `P_t=P_0*e^(kt) ; k and t` are growth

rate and time in years.

Population in 1998 is `P_5=5600 ; t= 5`

`5600 = 4420*e^(5k) or e^(5k)= 5600/4420`

Taking natural log on both sides we get,

`5*k* ln e= ln(280/221)~~ 0.2366 ; (ln e=1)` or

`k=~~ 0.2366/5~~~~0.0473`

The growth function is `P_t=4420*e^(0.0473t)`

Population in `2000 = ? ; t=2000-1993-7` years

`P_t=4420*e^(0.0473*7)~~ 6155.94`

Population in `2000` was `6156` [Ans]

Answer 2

The population in `2000` would be `6072`

Explanation:

Using Arithmetic Progression..

Let;

First Term `T_1 = a = 4420`

`1993, color(white)x 1994, color(white)x 1995, color(white)x 1996, color(white)x 1997, color(white)x 1998, color(white)x 1999, color(white)x 2000`

`color(white)x darr color(white)(xxx) darr color(white)(xxx) darr color(white)(xxx) darr color(white)(xxx) darr color(white)(xxx) darr color(white)(xxx) darr color(white)(xxx) darr`

`color(white)xT_1, color(white)(xxx) T_2, color(white)(xx) T_3, color(white)(xxx) T_4, color(white)(xx) T_5, color(white)(x,x) T_6, color(white)(x,x) T_7, color(white)(x,x) T_8`

Therefore;

`T_6 -> 1998`

Hence the Sixth Term is `5600`

Now to get the common difference..

Recall;

`T_6 = a + 5d = 5600`

`4420 + 5d = 5600`

`5d = 5600 - 4420`

`5d = 1180`

`d = 1180/5`

`d = 236`

Hence;

The common difference is `236`

From the above structure..

`T_8 -> 2000`

Hence, the Eight Term is `2000`

Recall;

`T_8 = a + 7d`

`T_8 = 4420 + 7(236)`

`T_8 = 4420 + 1652`

`T_8 = 6072`

Therefore, the population in `2000` would be `6072`