Question

Help Save the Lake!
A fisherman decided to stock his favorite lake with a few fish that are not native to the region. The fish he introduces have no natural predators in the lake and they experience a large exponential growth. The introduction of an invasive species can be very detrimental to an ecosystem and there have already been noted effects. The National Fish and Wildlife Foundation has asked for your help.

Answer 1

@amoodarya

Answer 2

idk on this

Answer 3

its ok Alex :)

Answer 4

A general exponential function has the form: y = a * b^x
where a is the initial value when x = 0 and b is the base.
If b > 1, then it will be an exponential growth.
If 0 < b < 1, then it will be an exponential decay.

Here they want growth, so we have to choose b > 1 and a has to be the initial number of fishes introduced.

1) Fill the table with data. For example,

Month (x) Population P(x)
0 100
1 110
2 121

Answer 5

ok

Answer 6

2) Derive an exponential function.
Let P(x) = a * b^x
Put the values from the table created in (1) into the equation P(x) = a * b^x:
when x = 0, P = 100: 100 = a * b^0 = a
a = 100
P(x) = 100 * b^x
Put the second table entry. When x = 1, y = 110
110 = 100 * b^1
b = 110/100 = 1.1
So the equation is: P(x) = 100(1.1)^x

Answer 7

3) Explaining the variables in the function: P(x) = 100(1.1)^x
x is the month
P(x) is the fish population when month is x.

Answer 8

So for my table I put 80, 100, 120 and 140 and then for #2 Let P(x)= a*b^x so x=0, P=80

Answer 9

80=a*b^0=a
a=80

Answer 10

The data you have chosen for the table is an arithmetic progression. It is increasing but it is not increasing exponentially. Your data is increasing linearly.

Instead make it a geometric progression.

Multiply by a factor, say 1.1. 80, 88, etc..

Answer 11

ok I'll do that one, 1.1 so it would be 0, 80. 1, 88, 2, 96.8 and 3 106.48

Answer 12

Can't have fractional fishes! :)

Answer 13

lol sorry, so it would be 97, and 107? Is that ok?

Answer 14

97 and 106.

Answer 15

oh ok. So it would be x=0, P=80, 80=a*b^0=a a=80 and then for the second one it would be x=1, y=97 97=80*b^1 b=97/80 which is 1.2

Answer 16

when x = 1, P = 88

Answer 17

oh yeah thats what I meant, sorry!! :/

Answer 18

97 is the 3 one.

Answer 19

1)
Month (x) Population P(x)
0 80
1 88
2 97
3 106

2) P(x) = a * b^x
when x = 0, P = 80. 80 = a * 1. a = 80
P(x) = 80 * b^x
when x = 1, P = 88. 88 = 80 * b^1. b = 88/80 = 1.1
P(x) = 80(1.1)^x

Answer 20

OMG ur like the best! And then number 3 is P(x)=80(1.1)^x and x is the month so it would be P(x)=80(1.1)^1 for the 1st one?

Answer 21

For 3) Explain all numbers and variables in the exponential function.
P(x) = 80(1.1)^x
80 is the number of fishes initially introduced.
1.1 is the base of the exponential function. It can be written as (1 + 0.1) where 0.1 represents the rate of growth. 0.1 or 10% is the rate of growth. Each month the fish population increases by 10%.
x is the month
P(x) is the fish population in the x-th month.

Answer 22

oh ok I see now, and then for the 4th one it says "4. The current exponential function P(x) measures the fish population’s growth monthly. Using complete sentences explain how to find the rate of growth every week." So for The rate of growth would be determined by the equation P(x)= 80(1.1)^x and that x would be the number of weeks in a month which would be 4. Well really it would be 4.34 but we cant have that so it would just be 4.

Answer 23

P(x) = 80(1.1)^x
We have constructed an exponential function where the exponent x should be in months.
In 4) they want to find the fish population by week and not month.
Let w represent the number of weeks.
There are 12 months in a year and there are 52 weeks in a year.
52 weeks = 12 months
w weeks = w * 12 / 52 = 3/13 * w months
Replace x in function with 3/13w months
P(w) = 80(1.1)^(3/13*w)

Answer 24

how do you get 13?

Answer 25

12/52 = 6/26 = 3/13

Answer 26

oh ok, simplified.

Answer 27

yeah.

Answer 28

So i was totally wrong on that one? :/

Answer 29

It is a common mistake. Replace x months by 4x weeks. But here the exponential function was constructed assuming the exponent was in months and not weeks. So "w" weeks will have to be converted to months by multiplying by 3/13. So now, even though we have the variable w in the exponent, 3/13*w is still in months.

Answer 30

oh my gosh, things are actually making sense now!! I have 2 more problems within this one,are you still able to help me with them?

Answer 31

5) When does the fish population reach 500.
Let us use the original month equation we constructed in 2)
P(x) = 80(1.1)^x
Set P = 500 and solve for x.
500 = 80(1.1)^x
divide by 80
500/80 = 1.1^x
6.25 = 1.1^x
Take logarithm on both sides
log(6.25) = log( 1.1)^x = x * log(1.1)
x = log(6.25) / log(1.1) = 19.23 months.

The fish population will reach 500 after a little over 19 months.

Answer 32

6) Convert the exponential function to a logarithmic function.
P(x) = 80(1.1)^x
Take logarithm to the base 10 on both sides:
log(P) = log( 80 * 1.1^x ) = log(80) + log( 1.1^x ) = log(80) + xlog(1.1)

log(P) = log(80) + log(1.1) * x is the logarithmic equation where the logs are with respect to base 10.

Answer 33

OMG you're such a lifesaver!!!!!!!!!!!!! if i could give u more than 1 medal at a time I would!! thank you soooooooooooo much!!!

Answer 34

You are welcome. Glad to be able to help. :)