First of all, we have to check out what the yearly decrease really amounted to. Then we'll know if the equation we are using is correct.
If there is a 26% decrease, then there's 74% of the population that remains. So after 1 year, 74% of 426000 people still live in this community. How many people are still there?
P = 426000 * (74/100) ^1 P = 315240 are still left in our community. What about the second year? This is really critical, because if we make a mistake in our theory, it will show up here.
P2 = 315240 * (74/100) P2 = 233277 People still live in this community.
Could we have found P2 directly? Yes. P2 = 426000 * 0.74^2 P2 = 233277 is the answer. So now we know what the answer to the question must be
year one:315240 is the population in the first year year two:233277 is the population in the second year
Now we're ready to begin answering your question. If we start out with P =426000 ((1 - 0.26)^1/12)^(12t) we should be able to get a monthly drop for a yearly %. You better read this over a couple of times to get the idea.
So P = [426000*(.74)^(1/12)]^12t is what you do. By the law of powers, you have done nothing to the actual equation, because the 12s cancel.
Put this in your calculator .74^(1/12) = 0.97522 Ans our equation becomes P = 426000 * 0.97522^(12t) where t is in years.
To check the equation, we will let t = 1 P = 426000 * 0.97522^12 P = 315240
But the acid test is year two. P = 426000 * 0.97522^24 P = 233277 which checks with doing it one step at a time using 0.74
So the answer is P = a * 0.97522^12t b = 0.97522 c = 12*t
