Question

An endangered species of fish has a population that is decreasing exponentially (A = ae kt). The population 9 years ago was 1700. Today, only 1000 of the fish are alive. Once the population drops below 100, the situation will be irreversible. When will this happen, according to the model? (Round to the nearest whole year.)

Answer 1

So we start with:\[
A=ae^{kt}
\]

Answer 2

A=1000e^ (i get stuck here)

Answer 3

Okay so you have: \[
A = 1000e^{kt}
\]

Answer 4

That was using the fact `Today, only 1000 of the fish are alive`

Answer 5

Next you want to use the fact: `The population 9 years ago was 1700.`

Answer 6

drops by a factor of \(\frac{10}{17}\) every 9 years
if you don't want to go through all that mess, use
\[\large 1000\times \left(\frac{10}{17}\right)^{\frac{t}{9}}\]

Answer 7

This is saying \(t=-9 , A=1700\)

Answer 8

So \[
1700=1000e^{k(-9)}
\]

Answer 9

We can solve for \(k\) now: \[
1.7 = e^{-9k}\implies \ln(1.7) =-9k \implies k = -\frac{\ln(1.7)}{9}
\]

Answer 10

Now we get: \[
A = 1000e^{-\ln(1.7)t/9}
\]

Answer 11

to get rid of the \(\ln\) we can use that: \[
e^{\ln(1.7)} = 1.7
\]This leads us to: \[
A = 1000 (1.7)^{-t/9}
\]

Answer 12

how do I get rid of the t? and thanks so much, this is very helpful.

Answer 13

We don't want to get rid of \(t\). We want to solve for it.

Answer 14

We use the final question to do so:
`Once the population drops below 100, the situation will be irreversible. When will this happen, according to the model`

Answer 15

This means \(A<100\).

Answer 16

So we want: \[
1000(1.7)^{-t/9} < 100
\]We need to solve it. Our answer should be along the lines of:\[
t> \ldots
\]Because we know that it's from some point in time and all times beyond that into the future.

Answer 17

What Im not getting is the t/9. i do not know what to do with it.

Answer 18

im sorry I am asking so much questions, I am not that good at math.

Answer 19

Okay, first think we do is divide simplify a bit: \[
(1.7)^{-t/9} < 0.1
\]

Answer 20

Now we need to remember that we have: \[
(1.7)^{-t/9} = (1.7)^{t(-1/9)}
\]

Answer 21

We can get rid of this \(-1/9\) by using the exponent laws: \[
(x^{-1/9})^b = x^{-1/9 \times b}
\]

Answer 22

We want \[
-1/9\times b = 1
\]So solving for \(b\) gives us \(b=-9\).

Answer 23

So we take both sides to the power of \(-9\): \[
[(1.7)^{t(-1/9)}]^{-9} < (0.1)^{-9}
\]This simplifies to: \[
(1.7)^t < (0.1)^{-9}
\]

Answer 24

This next part is a bit tricky, but we have to use logarithms.

Answer 25

Hold on, when I raised to the power of \(-9\), the inequality should have flipped.

Answer 26

\[
(1.7)^t > (0.1)^{-9}
\]Raising to negative powers flips inequalities.

Answer 27

\[
\ln[(1.7)^t] > \ln[(0.1)^{-9}] \implies t\ln(1.7) > -9\ln(0.1)\implies t> -9\frac{\ln(0.1)}{\ln(1.7)}
\]

Answer 28

i got 39 years. I think thats wrong.

Answer 29

http://www.wolframalpha.com/input/?i=1000%281.7%29%5E%28-t%2F9%29+%3C+100%2C+solve+for+t

Answer 30

Their solution is saying \(t>39\)

Answer 31

with your help, I was right with the 39 yearss?

Answer 32

I believe so.

Answer 33

thanks alot:-)