Exponential model

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lucky13

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During 1991, 150,000 people visited a particular amusement park. During 1995, the number had grown to 350,000.
What is the exponential model for this data?
Letting N(t) represent the number of visitors in any year and t represents the number of years since 1991
 
lucky13 said:
During 1991, 150,000 people visited a particular amusement park. During 1995, the number had grown to 350,000.
What is the exponential model for this data?
Letting N(t) represent the number of visitors in any year and t represents the number of years since 1991
Click to expand...

You are given a good start - by defining the variables - now what?

Please share with us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

What does an exponential model look like?
 
N(t) = No * e[sup:2st8tjfi]kt[/sup:2st8tjfi]

at t = 0 (1991) we have N = No = 150000

Now we have

N(t) = 150000 * e[sup:2st8tjfi]kt[/sup:2st8tjfi]

we know at t = 4 (1995) we have N(t) = 350000

350000 = 150000 * e[sup:2st8tjfi]k*4[/sup:2st8tjfi]

Now solve for 'k'
 
350000/150000(ln) =(ln)e^k4 << What does that mean?

=1.977=k4 <<< How did you get that?

=1.977/4=k

...That doesnt look right.
 
I multiplied both sides by ln to get the exponent down.

The number shound be .847 (ln(350000/150000), not 1.977
 
\(\displaystyle A(t) \ = \ Ce^{kt}, \ exponential \ growth \ model\)

\(\displaystyle A(0) \ = \ 150,000 \ = \ Ce^{k(0)}, \ \implies \ C \ = \ 150,000\)

\(\displaystyle A(t) \ = \ 150,000e^{kt}, \ A(4) \ = \ 350,000 \ = \ 150,000e^{4k} \ \implies \ k \ = \ ln(7/3)^{1/4}\)

\(\displaystyle Hence, \ A(t) \ = \ 150,000e^{ln(7/3)^{t/4}} \ = \ 150,000(7/3)^{t/4}, \ t(0) \ = \ 1991, \ t(4) \ = \ 1995.\)
 
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