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.
Best Answer
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 $$t=\frac{20 \log (3)}{\log (2)} \simeq 31.6993$$