How can algebra and logarithms be used to solve a penguin population problem?

[NachoKing]

[I have here a question from one of my many revision sheet in preperation for my exams.
The problem I'm having is with this parcticular question.[/SIZE]

A 10 year research program is being carried out on two penguin rookery populations in Tasmania. Rookery A had 2000 penguins at the start of the study and the population has been decreasing by 5% per year. Rookery B had 3000 penguins initially and this population has been decreasing by 10% per year.

a) Calculate how long it will take for the two populations to be the same.
b) Find this actual figure.

State any assumptions and use algebra, logarithms and other mathematics means (not a graphics calculator) to assist in your solution.

I assumed it was to include growth and decay, so i started trying some things with that and got stuck very quickly.

Any help is greatly appreciated.

 

[lanedance]

you're correct its to do with exponetial decay... show your work & where you're stuck

if the population at time t is N(t), you should start by setting a a simple first order DE, for the rate of change of the population dN/dt, in terms of N(t)

 

[jahaddow]

I have the same problem working out this, I don't know where to start, Could someone please help.

 

[NachoKing]

The first thing i attempted to do was find the decay factor using

f(x) = AxB^X

so for rookery A (2000 penguin, at a decay rate of 5% each year.)

But i am unsure on how to apply this to the formula.

 

[xX-Cyanide-Xx]

Hey, this question looks like a wopa, can someone pleaseshow the working out?

 

[HallsofIvy]

The first thing i attempted to do was find the decay factor using

f(x) = AxB^X

so for rookery A (2000 penguin, at a decay rate of 5% each year.)

But i am unsure on how to apply this to the formula.

X is number of years? Since any number to the 0 power is 1, when X= 0, f(0)=A= 2000.

Since this is a calculus problem, you can try using the derivative to get the decay rate equal to .05.

By the way, when you are using "X" as a variable, it is not a good idea to use "x" to represent multiplication. Use parentheses: f(x)= A(B^X).

 

[lanedance]

ok with Rookery A, you know
$$N_A(0) = 2000$$
$$N_A(1) = 0.95 N_A(0)$$

assume an exponetial decay [itex] N_A(t) = A_0 e^{-ct}[/tex], and use the constraints above to solve for the constants

 

[jahaddow]

Assuming that the decay rate for each population does not change before the populations are equal:

(a) A = 2000(1 - 0.05)^n

So A = 2000 * 0.95^n [eqn 1]

B = 3000 * 0.90^n

Now pops are the same when A = B

so 2000 * 0.95^n = 3000 * 0.90^n

0.95^n / 0.90^n = 3/2

Taking logs to base 10 of both sides:

log (0.95^n / 0.90^n) = log (3/2)

log 0.95^n - log 0.90^n = log (3/2)

n log 0.95 - n log 0.90 = log (3/2)

n(log 0.95 - log 0.90) = log (3/2)

n = log (3/2) / (log 0.95 - log 0.90)

n = 7.5

So it will take 7.5 years (to 1 dec pl) for the two populations to be the same
(b) Subs n = 7.5 into [eqn 1] → A = 2000 * 0.95^7.5

A = 1361

So the populations will both be 1361 penguins

 

[NachoKing]

Alrite, cheers jhaddow. Now it makes sense to me. That's all i need to see. thankyou
thank you thankyou.

 

[lanedance]

one step tha might help is as follows
0.95^n / 0.90^n = 3/2

then
(0.95/ 0.90)^n = 3/2

log of boths sides
n*log(0.95/ 0.90) = log(3/2)