this is one of my exponential logarithmic function can someone help me

Question

The population of a city has been growing exponentially. The data table shows the city population in thousands over a 40 year period.

 

Year	1970	1980	1990	2000	2010
Population (thousands)	219.5	268.0	326.5	397.5	485.0

 

a) Determine the exponential equation that models this data. [2]

b) Use your equation (or curve of best fit) to estimate the population in 2015, to the nearest hundred people. [2] Show your work.

Answer 1

Let p(y)=ae^by+c. When y=1970 we want p(1970)=219.5, so if a=219.5 then 1970b+c=0, c=-1970b.

So p(y)=219.5e^by-1970b or 219.5e^b(y-1970).

Now we have to find b. Let’s take the population of two consecutive decades:

y=1970, p(1970)=219.5.

y=1980, p(1980)=268.0.

p(1980)/p(1970)=e^1980b+c/e^1970b+c=e^10b=268/219.5=1.2210.

10b=ln(1.221)=0.2, so b=0.02.

We have taken the ratio of the populations of 1970 and 1980. We need to consider populations from other years to be sure there’s constant growth.

So 326.5/268=1.2183, 397.5/326.5=1.2175, 485/397.5=1.2201.

They’re all fairly close to one another, so b=0.02 is a fair estimate.

We could simply take logs of each population and compare them:

ln(219.5)=5.3914, ln(268)=5.5910, ln(326.5)=5.7884, ln(397.5)=5.9852, ln(485)=6.1841.

There’s a common difference of about 0.20, but if we take the average difference we get 0.1982. ln(1.2192)=0.1982=10b, b=0.01982, so the average ratio of the decade populations is 1.2192. Let’s see what b=0.02 predicts (to the nearest 0.5):

1970 219.5

1980 268.0

1990 327.5

2000 400.0

2010 488.5

2015 540.0

Compare with b=0.01982:

1970 219.5

1980 267.5

1990 326.5

2000 398.0

2010 485.0

2015 535.5

The second model fits better than the first, so:

(a) p(y)=219.5e^{0.01982(y-1970)} is a good model.

(b) prediction for 2015 using model in (a) is population (thousands) = 535.5

Another way of looking at this problem is to consider a geometric series where the first term is a, the next term is ar, then ar², ar³, etc. So a would be 219.5 and r would be the average increasing rate. The data shows decades and we are asked to predict the population for 2015, which is not a decade. So r has to be the rate of increase of no more than 5 years. If we make r the annual increase, then the series would be a, ar¹⁰, ar²⁰, ar³⁰, etc. But we need to relate these exponents to the actual year, so we get ar^y-1970. We already know that r¹⁰=1.2192. From this 10ln(r)=ln(1.2192)=0.1982, so ln(r)=0.01982, making r=1.02 (about 2% per year). So p(y)=219.5(1.02)^y-1970. Lets try it out:

1970 219.5

1980 267.5

1990 326.0

2000 397.5

2010 484.5

2015 535.0

(To the nearest 0.5 thousands, that is, to the nearest 500). The model fits well.