Formula A=AoE^tk?

Question

I have a question that askes:
A bird population is going extinct exponentially. 5 years ago there were 1400 birds, now there are 1000. Using A=Aoe^tk when will the population hit 100?
If someone could set up the formula for me I could solve it, I just don't know where to start.
Thanks

Answer 1

The first answer is on the right path, but leaves out a couple of steps. Your equation has 2 constants you need to determine, A0 and k. The value of A at two time points gives you 2 equations. If time 0 starts 5 years ago, then A0=1400 using t=0. And A=1000=1400e^(5k) will give you k. Now you can use your equation to find t when A=100.

Answer 2

they give you two points. let t = 0 since 5 years ago.
(0,1400) is one of the point

5 years later, the population is 1000
(5,1000) is the other point

Ao is the intial popopulation

A = 1400e^(kt)

now use the other point
1000 = 1400e^(k*5)

solve for k
5/7 = e^(5k)
ln(5/7) = 5k
k = ln(5/7)/5
k =~ -0.0673

A = 1400e^(-0.0673t)

now just plug 100 for A and solve for t. Remember the t value in this prolem is HOW LONG it takes the poplulation to reach 100. Not when. Since the question didn't give a specific year, we can't determine when.

Answer 3

Assume that the function is the one given:

A = A0* e^(kt)

Let 5 years ago be t = 0 ---> A = A0 =1400 birds

Now at t = 5 years A = 1000 = A0*e^(5k)

1000 = 1400*e^(5k)

1/1.4 = e^(5k) --->k =(1/5 )* ln(1/1.4) = -0.2*ln(1.4) = -0.0673 years^-1

SO now we know A0 and k. To find t when A = 100;

A =A0e^(kt) --->kt = ln(A/A0) ---> t = 1/k*ln(A/A0)

t = -1/(0.0673)*ln(100/1400) = 39.22 years

Answer 4

You need to solve for k first.

1000 = 1400*e^(5*k)

k = ln(1000/1400)/5

Just plug in 100 for 1000 and solve for t

t = ln(100/1400)/k

Plug and chug.

Answer 5

1,000 = 1,400e^5k
5/7 = e^5k
(ln (5/7)) / 5 = k

k = -0.06729444732...

100 = 1,400e^(-0.06729444732...t)
1/14 = e^(-0.06729444732...t)
(ln (1/14)) / [(ln (5/7)) / 5] = t
t ~ 39.2 years later from when there were 1,400 birds