Exponential problem

[EvoN1020v]

Hello guys. I was doing my math review in preparation for my Advanced Trigonometry exam, and I encountered this question with a difficulty.

 

When there were 1000 hares ([math]N_h = 1000e^{kt}[/math]), the number of foxes was increasing so that [math]N_f = 74e^{0.08t}[/math]. How many years will it take for the number of foxes to reach 10% of the number of hares?

 

I know I have to use the Natural logarithms to solve this question, but I'm not sure what I should do? If you guys can give me hints, that would be great, or I'll have to ask my teacher when I go back to school.

[Klaynos]

Well if you equate one of them to 10% of the other one and then use natural log, you should see where you've got to go from there.

[EvoN1020v]

Yeah? I need to figure out how many years to get the population of foxes to be 10% of the population of hares. Do I put the 2 equations together and figure out the t? (t=years)

[Klaynos]

yes

[EvoN1020v]

I know, but any suggestion of how to do that?

[Klaynos]

N_f=0.1*N_h

 

Substitute your values for the two N's into that and Ln both sides.

[EvoN1020v]

This is frustating. The [math]N_f[/math] seems to be missing the value for k. I don't know much the hares increase or decrease?? Or the foxes to be assumed 10% of the original amount of hares??

[timo]

Either you didn´t post the whole question or k is simply a free parameter. Just solve for the time t, then.

[EvoN1020v]

If the foxes' population to be 10% of the original population of hares of 1000, then the population of foxes to be 100.

 

[math]N_f = (0.1)(N_h)[/math]

[math]N_f = (0.1)(1000)[/math]

[math]N_f = 100[/math]

 

Then I put the 100 into the foxes' population forumla:

 

[math]100 = 74e^{0.08t}[/math]

 

[math]\frac{100}{74} = \frac{74e^{0.08t}}{74}[/math]

 

[math]ln1.351351351 = lne^{0.08t}[/math]

 

[math]ln1.351351351 = \frac{0.08tlne}{0.08lne}[/math]

 

[math]t = \frac{ln1.351351351}{0.08lne}[/math]

 

[math]t = 3.76381366[/math]

 

This answer to be 10% of hares' original population. If I am mistaken, please let me know. Thanks.

[MattC]

The k represents a constant, so treat it as one. That's my suggestion.

 

1000e^kt=740e^0.08t

 

You can solve this for "t", right? I can, and here's my hint: the answer has in the denominator the variable "k."

[timo]

I don´t think you should have posted the solution Matt. There´s a an error (probably a typo) in your calculation anyways, so perhaps you should consider removing it completely. It´s pretty naive to think people won´t look at solutions when they are given instead of really doing all of the calculation themselves. The question isn´t that hard, anyways and I think the confusion only stems from the free parameter you called a constant (which in fact for calculation terms might be the better way to think of it).

[MattC]

oh. What error was it? I'm a little tipsy, to be honest. Had a long day at work and I'm having a beer and reading scienceforums to relax.

[EvoN1020v]

Nope. I cannot solve for "t" when I have no value for "k" on the hares' equation.

[MattC]

He did say it was in preparation for an exam, but I agree, it's not wise to help others too much with their homework. That would attract the wrong crowd.

[timo]

Judging from how you solved the problem with the 10% of the original population I´d say you are missing an important equation for logarithms: ln(x*y) = ln(x)+ln(y).

As a little style-advice: Don´t write things like ln(e), write 1 directly (or skip it if it´s a factor). It makes your calculation much more readable.

 

@MattC: I´m pretty sure it was a typo. But looking at your result you could have figured out that it couldn´t be correct. I´ll send you a PM. It doesn´t matter whether it is in preparation for an exam or homework. The point is that you don´t learn as much from seeing other people´s solutions as from solving it yourself. And I´m pretty convinced he can solve it himself - there´s not so much possibilities to take logarithms here, anyways.

[MattC]

I'm suprised there was only one typo in my solution. I would have removed it regardless, though - better to just give a hint, like you did. Ah, speaking of typos ... may want to edit that equation fast!