# study question, Alg II

• Oct 6th 2008, 01:45 PM
Toddry
study question, Alg II
I am having major issues with this problem, even with starting it. I have a test coming up and this is an example. I actualy have 3 questions but this is the first.

The population of a midwestern city follows the exponential law.
a.) If N is the population of the city and t is the time in years, express N as a function of t.

b.) If the population decreased from 800,000 to 700,000 from 2003 to 2005, what will be population in 2007.
• Oct 6th 2008, 05:24 PM
ticbol
Quote:

Originally Posted by Toddry
I am having major issues with this problem, even with starting it. I have a test coming up and this is an example. I actualy have 3 questions but this is the first.

The population of a midwestern city follows the exponential law.
a.) If N is the population of the city and t is the time in years, express N as a function of t.

b.) If the population decreased from 800,000 to 700,000 from 2003 to 2005, what will be population in 2007.

I believe the population exponential rule is P = (Po)e^(kt)
where Po = initial P
k = any rational number
t = time

So from 2003 to 2005,
t = 2 years
Po = 800,000
P = 700,000
k = ?

700,000 = 800,000e^(k*2)
e^(2k) = 700,000 /800,000 = 7/8
2k *ln(e) = ln(7/8)
k = ln(7/8) /2
k = -0.066766

Then, from 2003 to 2007,
t = 4 years
Po = 800,000
P =?
k = -0.066766

Can you solve for P now?
You should get P = 612,500
• Oct 7th 2008, 05:13 AM
Toddry
Quote:

Originally Posted by ticbol
I believe the population exponential rule is P = (Po)e^(kt)
where Po = initial P
k = any rational number
t = time

So from 2003 to 2005,
t = 2 years
Po = 800,000
P = 700,000
k = ?

700,000 = 800,000e^(k*2)
e^(2k) = 700,000 /800,000 = 7/8
2k *ln(e) = ln(7/8)
k = ln(7/8) /2
k = -0.066766

Then, from 2003 to 2007,
t = 4 years
Po = 800,000
P =?
k = -0.066766

Can you solve for P now?
You should get P = 612,500

I am going to check it now. Thank you much. I am coming up with a lot of the same. Sometimes its hard to start these.