# Thread: WORD

1. ## WORD

I know you are suppose to use y(t) = e^(kt) but I don't know where to go. Please help me out guys. Thanks In 1970 the population of a town was 21,000 and by 1980 that number decreased to 20,000. If the population decreases exponentially, then
a. find the population in the year 2000
b. find the population in the year 1960

2. This is a hard one but I think you would use the years as t and continue from there

3. Hello, bigton!

A piece that formula is missing . . .

Use: .$\displaystyle y(t) \:= \:{\color{red}a}e^{kt}$

In 1970 the population of a town was 21,000
and by 1980 that number decreased to 20,000.
If the population decreases exponentially, then

a) find the population in the year 2000

Let 1970 be $\displaystyle t = 0.$

We are told that: .$\displaystyle y(0) \:=\:21,\!000$

. . We have: .$\displaystyle 21,\!000 \:=\:ae^0 \quad\Rightarrow\quad a \:=\:21,\!000$

The function is: .$\displaystyle y(t) \;=\;21,\!000e^{kt}$

We are told that: .$\displaystyle y(10) \:=\:20,\!000$

. . $\displaystyle 20,\!000 \:=\:21,\!000e^{10k} \quad\Rightarrow\quad e^{10k} \:=\:\tfrac{20}{21} \quad\Rightarrow\quad 10k \:=\:\ln\left(\tfrac{20}{21}\right)$

. . $\displaystyle k \:=\:\tfrac{1}{10}\ln(\tfrac{20}{21}) \:=\:-0.004879016 \;\approx\;-0.005$

The function is: .$\displaystyle y(t) \;=\;21,\!000e^{-0.005t}$

In the year 2000, $\displaystyle t = 30\!:\;\;y(30) \;=\;21,\!000e^{(\text{-}0.005)(30)} \;=\; 18074.8675 \;\approx\;18,\!075$

b) find the population in the year 1960

In the year 1960, $\displaystyle t = \text{-}10$

$\displaystyle y(\text{-}10) \;=\;21,\!000e^{(\text{-}0.05)(\text{-}10)} \;=\;21,\!000e^{0.05} \;=\;22076.69302 \;\approx\;22,\!077$

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