Homework Help Forum: Math

Posted by SC on Thursday, November 5, 2009 at 8:17pm.

The population of termites and spiders in a certain house are growing exponentially. the house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders/ Hows long in days does it take the population of spiders to triple? What I done so far. TERMITES y(not)=100 t=4 y=200 200=100b^4 b=2^(1/4) T(t)=100(1.189207)^t for the spiders.S(t).. i tried plugging t=3 and t=8and divided by two and four to the T(x) equation.... and for two numbers 84.089 and 99.99..... but I think Im doing it wrong or is it T(t) = 2(S(t))?

• Math - TeacherJim, Thursday, November 5, 2009 at 8:47pm

  I agree with your first part: b=2^(1/4) = 1.189207. Now, after 3 days, according to the question, we must have 84.089 spiders - wonder what .089 of a spider looks like? - and after 8 days, we have 100. If S0 is the number of spiders on day zero, and s is the multiplication rate of spiders, then S0 * s^3 = 84.089 S0 * s^8 = 100 From which we can quickly get s^5. And then we get s. And from there, there's not much left to do, I think.

  
  

• Math - SC, Thursday, November 5, 2009 at 8:57pm

  I made a mistake. after eight days , its 199.99 spiders. I still don't understand how would we get s?

  
  

• Math - TeacherJim, Thursday, November 5, 2009 at 9:07pm

  You didn't make a mistake: "Eight days after moving in, there were four times as many termites as spiders" But 100 * (2^.25)^8 = 400 (termites) So if there were 400 termites, there were 100 spiders. So how do we get s? S0 * s^3 = 84.089 S0 * s^8 = 100 (S0 * s^8) / (S0 * s^3) = s^5 = 100 / 84.089 s = (100 / 84.089)^(1/5)

  
  

• Math - SC, Thursday, November 5, 2009 at 9:10pm

  Ohh all i needed to do was divide! heh thanks a lot!

  
  

• Math - TeacherJim, Thursday, November 5, 2009 at 9:13pm

  You're welcome!

  
  

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