Question

In 2000 the population of Austraiia was 19.0 M, in 2010 it was 22.6 M. The exponential model
P=P_0 e^kt
describes the Australian population, P, in millions, t years after 2000.

Answer 1

Find the growth rate, k, to 5 decimal places.

Answer 2

t is the number of years since 2000. i.e. for year 2000, t=0
put in t=0 and P= 19.0
to create one equation

then do the same for t= 10 (meaning 2010) and P= 22.6

what two equations do you get ?

Answer 3

\[19.0=19.0_0 e ^{k+0}\]

Answer 4

almost. the \( 19.0_0 \) should stay \(P_0\)
also, it is k*t not k+t

Answer 5

oh yea my bad with the k+0 thing
why does it stay as \[p _{0}\]

Answer 6

P0 and k are the unknowns.
However, we are given P (population) and time t

that means we can form 2 equations (with 2 unknowns)

can you write the 2 equations?

Answer 7

Well I'm assuming that \[19.0=P _{0} e ^{k*0}\] is one but im not sure of the other @phi

Answer 8

the other one has a different P and a different t
in 2010 it was 22.6 M

t is the years since 2000

Answer 9

oh alright the other one is

Answer 10

\[22.6=P _{0} e ^{k*10}\]

Answer 11

ok. notice in the first one you have an exponent k*0 which is just 0
also e^0 is 1 (anything to the 0 power is 1)
so the first equation simplifies to
19.0= P0

that means we have found P0
use that value in the 2nd equation

Answer 12

22.6=19.0^k(10)

Answer 13

ok, except you left out the e
\[ 22.6= 19.0 e^{10k} \]

Answer 14

now divide both sides by 19.0
\[ \frac{22.6}{19.0} = e^{10k} \]

now take the "natural log" of both sides.
can you do that ?

Answer 15

\[\frac{ \ln(22.6) }{ \ln(19.0) }=10?\]

Answer 16

10k*

Answer 17

you take the ln of both sides like this
\[ \ln\left( \frac{22.6}{19.0}\right) = \ln \left( e^{kt}\right) \]

Answer 18

does the 10 and the k not come down then?

Answer 19

which is also
\[ \ln\left( \frac{22.6}{19.0}\right) = 10k \]

yes the 10k "comes down"
(It's important to memorize this property)

Answer 20

so then I just get the ln of that and solve

Answer 21

* kt should have been 10k

yes, we get (after dividing both sides by 10)
\[ 0.1\cdot \ln\left( \frac{22.6}{19.0}\right) = k \]

Answer 22

0.0173511?

Answer 23

@phi

Answer 24

yes. you can try it in the equation
\[ P = 19 e^{0.0173511 t} \]

Answer 25

the only tweak is they want k to five decimal places
0.01735