Modeling exponential growth quickly without doing too much math computation is what this lesson will show you.
In our previous lesson about what is exponential growth we modeled the following problem with y = 16,000,000(1.02)^{x}
The population of Florida was 16 million in 2000. Then every year, the population has grown by 2%
However, there is an easier and quicker way to get the same answer.
The only trick is to see that 100% + 2% = 102% = 1.02 instead of doing the complicated math computation in the lesson referenced above. Once you identify the growth factor, you can just model the problem as shown below.
Let x be the number of years since 2000
Let y be the population at various time
Let a be the initial population or 16,000,000
Let b be the growth factor, which is 1.02
Then using y = ab^{x}, we get y =16,000,000(1.02)^{x}
Other examples
Suppose that the cost of caring for patients in community hospitals has been increasing about 4% each year since 1996. If the cost was 998 dollars in 1996, model the cost of care since 1996 with an exponential function.
y = ab^{x}
Let x be the number of years since 1996
Let y be the cost of care in community hospitals for various times.
Let a be the initial cost in 1996 or 998 dollars
Let b be the growth factor, which is 100% + 4% = 104% = 1.04
Then y = 998(1.04)^{x}
Suppose you deposit 4000 dollars in a saving account paying 1.5% interest compounded annually or once per year on January 1, 1998. Model in years the amount of money you have in the saving account since 1998.
y = ab^{x}
Let x be the number of years since 1998
Let y be the amount of money in the saving account for various times.
Let a be the initial deposit in 1998 or 4000 dollars
Let b be the growth factor, which is 100% + 1.5% = 101.5% = 1.015
Then y = 4000(1.015)^{x}
Suppose instead that the interest was compounded quarterly instead of annually.
Find a model for this situation.
y = ab^{x}
x cannot be the number of years since 1998 anymore since interest is computed now every 3 months.
Let x be the number of interest periods. For example, 5 years after 1998, the number of interest periods is 5 times 4 = 20
Let y be the amount of money in the saving account for various times.
Let a be the initial deposit in 1998 or 4000 dollars
Let b be the growth factor, which is 100% + 1.5% / 4 = 1 + 0.00375 = 1.00375
Notice that we divide the interest into 4 parts since there are 4 interest periods in a year.
Then y = 4000(1.00375)^{x}
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