Question

Please help me with this

Answer 1

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

Answer 2

100%-26%
426000*(1-0.26)^t, yearly
(0.74^(1/12))=0.74^0.083 monthly rate of decrease
 Equation for calculation population
426000*(0.74^0.083)^12t

I think it should look like this 0.74^0.083 monthly rate of decrease