The populations of termites and spiders in a certain house are growing exponentially. The house contains 120 termites the day you move in. After four days, the house contains 210 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.

How long (in days) does it take the population of spiders to triple? (Round your answer to one decimal place.)

So this is everything I did: I got the termites exp equation as $y=120(1.1501633169)^x$ $So y=120(1.1501633169)^3$(days) $= 182.58276622 termites/2 = 91.29138311$ spiders For 8 days I got $91.875000002$ spiders.

So the growth rate would be $(91.875000002/91.29138311)^{8/3}$ to get $1.0171386889$

So for 3 times that population I got $ln(3)= t ln(1.0171386889) =64.6$

but it's wrong.

I don't know if I need to find the initial starting pt for the spiders bc I dont think it matters bc if I divide that into the 3 times amount I would just get 3. So I dont know where I went wrong on this problem.

  • $\begingroup$ Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer. $\endgroup$ – Claude Leibovici May 11 '14 at 4:58
  • $\begingroup$ No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it... $\endgroup$ – Sondra May 11 '14 at 7:42
  • $\begingroup$ There is another one with different numbers. I just cant figure out how to get the answer. Here it is: There are 100 termites when they move in, after 4 days there are 200. After 3 days there are 2 times as many termites as spiders. 8 days after there are 4 times as many termites as spiders. The answer for this one is 31.6993 days. $\endgroup$ – Sondra May 11 '14 at 7:54

Let me take the second problem you gave in your last comment.

First of all, we shall note $T(t)=T_0~a^t$ and $S(t)=S_0~b^t$ the populations of termites and spiders as a function of time.

Concerning the termites, the first informations you give translate to $$T(0)=100$$ $$T(4)=200$$ From these, $T_0=100$ and $a=\sqrt[4]{2}~~ (\simeq 1.18921)$.

Concerning the ratio between termites and spiders, the next informations translate to $$T(3)=2~S(3)$$ $$T(8)=4~S(8)$$ This write $$T_0~ a^3=2~ S_0~ b^3$$ $$T_0~ a^8=4~ S_0 ~b^8$$ and taking the ratio $$\frac{T_0~ a^8}{T_0~ a^3}=\frac{4~S_0~ b^8}{2~S_0~ b^3}$$ which simplifies to $a^5=2~b^5$ or $2^{\frac{5}{4}}=2~b^5$ or $2^{\frac{1}{4}}=b^5$ and finally $b=\sqrt[20]{2} ~~(\simeq 1.03526)$.

So now, the last question translates to $$S(t)=3~S(0)$$ As you noticed, we do not need to care about $S_0$ which eleminates from the equation. So what we need is to find $t$ such that $$\frac {S(t)} {S(0)}=\frac {S_0~b^t}{S_0~b^0}=b^t=3$$ Taking logarithms of both sides and replacing $b$ by its value, we arrive to $$t=\frac{20 \log (3)}{\log (2)} \simeq 31.6993$$

  • $\begingroup$ I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from? $\endgroup$ – Sondra May 12 '14 at 23:49
  • $\begingroup$ $b^t=(\sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=\frac{20 \log (3)}{\log (2)}$. $\endgroup$ – Claude Leibovici May 13 '14 at 5:09
  • $\begingroup$ yeah I got that but why is it to the 20th root? where did you get get the 20? $\endgroup$ – Sondra May 13 '14 at 5:10
  • $\begingroup$ Simplifying the results to get $b$. Logarithms again. $\endgroup$ – Claude Leibovici May 13 '14 at 5:12
  • $\begingroup$ okay yeah I do that I think because I take what the intial for spiders is and divide that into 84.09085 (spiders after 3 days). I then took the ln of those two divided that to get 1.109577 and since i took the ln of that it wouldve had to take the ln of b^3= 3 ln b to get 1.109577 divided by 3 = 0.3698593. so ln b= 0.369859 and then putting that to the e to give rid of ln i get 1.44753......where am I going wrong? $\endgroup$ – Sondra May 13 '14 at 22:00

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