Exponential word problem (i think)

Click For Summary
SUMMARY

The discussion revolves around solving an exponential decay problem related to population reduction. The initial population of 7500 is reduced by 25% annually, leading to a target population of 4000. The population can be modeled using the equation P(t) = C * e^(kt), where C is the initial population and k is the decay constant. The alternative approach using P = P_0 * (3/4)^n is also discussed, with the final solution indicating that it takes approximately 2.185 years for the population to reach 4000.

PREREQUISITES
  • Understanding of exponential functions and decay
  • Familiarity with natural logarithms and their properties
  • Basic knowledge of calculus concepts, particularly related to population modeling
  • Ability to manipulate equations and solve for unknown variables
NEXT STEPS
  • Study the derivation of exponential decay formulas in population dynamics
  • Learn how to apply natural logarithms to solve for unknowns in exponential equations
  • Explore real-world applications of exponential decay in ecology and conservation
  • Practice similar problems involving exponential growth and decay scenarios
USEFUL FOR

Students studying calculus, ecologists interested in population dynamics, and anyone looking to understand exponential decay models in real-world contexts.

Jacobpm64
Messages
235
Reaction score
0
I'm not sure if this is calculus, but it is like a review in my calculus class.

Suppose that in any given year, the population of a certain endangered species is reduced by 25%. If the population is now 7500, in how many years will the population be 4000?

I've known how to do this before. I just forgot how to set it up. I'm pretty sure it's with exponents though. Please help.
 
Physics news on Phys.org
If it appeared as a question for calculus review, then they probably want you to proceed like this: You know the population can be descrided by an exponential equation of the form
P(t)=Ce^{kt}
for unknown constants C and k. Knowing the initial population will allow you to solve for C. Knowing the population a year later will let you solve for k. Then solve for time in the equation by taking ln's. Plug in the population you want and get the time it takes.
 
so let's see..
P(t) = 4000
C = 7500
e = the constant 2.718 etc.
k = what's k? I'm not sure how to do it since it's decay...?
t = what I'm solving for

so what is k?
 
Jacobpm64 said:
P(t) = 4000
Not exactly:
P(t)=7500e^{kt}
And in one year the population will be reduced by 25%, so:
P(1)=7500-.25*7500=5625=7500e^{k*(1)}
So now you can solve for k.
 
It might be easier to start with

P = P_0 \left( \frac {3}{4} \right)^n

where n is the number if years.
 
Tide said:
It might be easier to start with
P = P_0 \left( \frac {3}{4} \right)^n
where n is the number if years.
Yeah, I was thinking that too, but when I did that in high school calc class I remember my teacher took off points because she said "This is calculus class, solve the problem with calculus". They are probably trying to teach exponentials with this problem, but its good to know how to do it both ways.
 
2.1850811 years?

I believe so.. confirmation?

and thanks a lot :)
 
Jacobpm64 said:
2.1850811 years?
I believe so.. confirmation?
and thanks a lot :)
That's right.
 
This was also posted under "homework".


1) Please do not post the same thing in two different places.

2) There is a "sticky" thread at the top of this area that says this is NOT the right place to post homework problems.
 
  • #10
yeah, sorry about that, it won't happen again.. i noticed that sticky after i posted this one.. so i reposted in homework.. all of my future ones will go in homework.. thanks for the help, nonetheless
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
12K
  • · Replies 17 ·
Replies
17
Views
12K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 51 ·
2
Replies
51
Views
14K