Question 635620
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You start with *[tex \LARGE y\ =\ P_oe^{ht}]


You know the starting population, *[tex \LARGE P_o] because that is given, and you know the time, *[tex \LARGE t] because it is 30 years from 1980 to 2010, and *[tex \LARGE e] is the constant base of the natural logs, so that is known as well.  What you don't know is *[tex \LARGE h], the <b><i>rate</i></b> of exponential growth.


Fortunately, we are given another data point, specifically the population in 1990, and using this, we can solve the equation for *[tex \LARGE h], when *[tex \LARGE y\ =\ 40,500], *[tex \LARGE P_o\ =\ 30,000], and *[tex \LARGE t\ =\ 10] since it is 10 years from 1980 to 1990.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 40,500\ =\ 30,000e^{10h}]


Divide by 30,000:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{10h}\ =\ 1.35]


Take the natural log of both sides.  Actually, you could take the log to any base, as long as it is the same base on both sides of the equation, but because the equation involves *[tex \LARGE e], the base of the natural logs, the arithmetic is going to be much simpler if you take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{10h}\right)\ =\ \ln(1.35)]


Now use *[tex \LARGE \log_b\left(a^n\right)\ =\ n\,\cdot\,\log_b(a)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10h\,\cdot\,\ln(e)\ =\ \ln(1.35)]


Then use *[tex \LARGE \log_b(b)\ =\ 1] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10h\ =\ ln(1.35)]


(See what I mean about simpler arithmetic?)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ \frac{\ln(1.35)}{10}]


Which the calculator says is close enough to 0.03 that the difference doesn't matter.


NOW, you go back to the original function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ P_oe^{ht}]


Plug in the numbers


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 30,000e^{0.03\,\cdot\,30}]


And do the arithmetic.  I'll let you run the calculator this time.  


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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