SOLUTION: Here is the problem I am working on: A bird species in danger of extinction has a population that is decreasing exponentially. Five years ago the population was at 1400 and toda

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Here is the problem I am working on: A bird species in danger of extinction has a population that is decreasing exponentially. Five years ago the population was at 1400 and toda      Log On


   

Question 395078: Here is the problem I am working on:
A bird species in danger of extinction has a population that is decreasing exponentially. Five years ago the population was at 1400 and today only 1000. Once the population drops below 100, the situation will be irreversible. When will this happen?
I will use the model A=Aoe^kt
I would appreciate any help! Thank You!
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
First we use general exponential model equation and the initial information to find the specific exponential equation for this problem. In the general equation, the A%5B0%5D represents "the A when t = 0. The earliest value for A we have is from five years ago. So A%5B0%5D = 1400. This makes the other data, 1000 birds (A = 1000) five years later (t = 5) another point that should fit this equation. We now know 3 of the 4 unknowns in the general model. We can use the equation to find the 4th unknown, k:
1000+=+1400%2Ae%5E%28k%285%29%29
First we divide by 1400:
1000%2F1400+=+e%5E%28k%285%29%29
which simplifies to
5%2F7+=+e%5E%285k%29
Next we find the natural log of each side:
ln%285%2F7%29+=+ln%28e%5E%285k%29%29
Then we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent of the argument out in front. (It is this property that is the very reason we use logarithms on equations where the variable is in an exponent!)
ln%285%2F7%29+=+%285k%29%2Aln%28e%29
By definition, ln(e) = 1 so this becomes:
ln%285%2F7%29+=+5k
Dividing by 5 we get:
ln%285%2F7%29%2F5+=+k
This is an exact expression for k. This makes the specific equation to model this problem:
A+=+1400%2Ae%5E%28%28ln%285%2F7%29%2F5%29t%29

For practical use we should replace the k with its decimal approximation (which we can find using our calculator):
A+=+1400%2Ae%5E%28%28-0.3364722366212129%2F5%29t%29
A+=+1400%2Ae%5E%28-0.0672944473242426t%29

We can now use this equation to find when the bird population drops below 100:
99+=+1400%2Ae%5E%28-0.0672944473242426t%29
We solve this for t, just like we solved the earlier equation for k:
99%2F1400+=+e%5E%28-0.0672944473242426t%29
ln%2899%2F1400%29+=+ln%28e%5E%28-0.0672944473242426t%29%29
ln%2899%2F1400%29+=+%28-0.0672944473242426t%29%2Aln%28e%29
ln%2899%2F1400%29+=+-0.0672944473242426t%29
ln%2899%2F1400%29%2F%28-0.0672944473242426%29+=+t%29
%28-2.6491076654687601%29%2F%28-0.0672944473242426%29+=+t%29
39.3659175578729793 = t
So the bird population will drop below 99 approximately 39.4 years after the t=0 time. Since the t=0 time was 5 years ago, the bird population will drop below 100 approximately 34.4 years from now.