Solving Exponential Function for termites vs spiders











up vote
1
down vote

favorite












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.










share|cite|improve this question
























  • Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
    – Claude Leibovici
    May 11 '14 at 4:58










  • No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
    – Sondra
    May 11 '14 at 7:42










  • 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.
    – Sondra
    May 11 '14 at 7:54

















up vote
1
down vote

favorite












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.










share|cite|improve this question
























  • Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
    – Claude Leibovici
    May 11 '14 at 4:58










  • No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
    – Sondra
    May 11 '14 at 7:42










  • 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.
    – Sondra
    May 11 '14 at 7:54















up vote
1
down vote

favorite









up vote
1
down vote

favorite











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.










share|cite|improve this question















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.







algebra-precalculus exponential-function word-problem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '17 at 7:26









Jon Rose

1053




1053










asked May 11 '14 at 1:23









Sondra

941110




941110












  • Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
    – Claude Leibovici
    May 11 '14 at 4:58










  • No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
    – Sondra
    May 11 '14 at 7:42










  • 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.
    – Sondra
    May 11 '14 at 7:54




















  • Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
    – Claude Leibovici
    May 11 '14 at 4:58










  • No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
    – Sondra
    May 11 '14 at 7:42










  • 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.
    – Sondra
    May 11 '14 at 7:54


















Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
– Claude Leibovici
May 11 '14 at 4:58




Are they given the answer ? I ask this question because I obtained something ridiculous and I prefer to check before giving an answer.
– Claude Leibovici
May 11 '14 at 4:58












No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
– Sondra
May 11 '14 at 7:42




No there is no answers given. Its an online assignment and I have so many tried before I get a zero on it...
– Sondra
May 11 '14 at 7:42












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.
– Sondra
May 11 '14 at 7:54






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.
– Sondra
May 11 '14 at 7:54












1 Answer
1






active

oldest

votes

















up vote
0
down vote













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 $$t=frac{20 log (3)}{log (2)} simeq 31.6993$$






share|cite|improve this answer























  • I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
    – Sondra
    May 12 '14 at 23:49










  • $b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
    – Claude Leibovici
    May 13 '14 at 5:09












  • yeah I got that but why is it to the 20th root? where did you get get the 20?
    – Sondra
    May 13 '14 at 5:10










  • Simplifying the results to get $b$. Logarithms again.
    – Claude Leibovici
    May 13 '14 at 5:12










  • 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?
    – Sondra
    May 13 '14 at 22:00











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f789788%2fsolving-exponential-function-for-termites-vs-spiders%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













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 $$t=frac{20 log (3)}{log (2)} simeq 31.6993$$






share|cite|improve this answer























  • I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
    – Sondra
    May 12 '14 at 23:49










  • $b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
    – Claude Leibovici
    May 13 '14 at 5:09












  • yeah I got that but why is it to the 20th root? where did you get get the 20?
    – Sondra
    May 13 '14 at 5:10










  • Simplifying the results to get $b$. Logarithms again.
    – Claude Leibovici
    May 13 '14 at 5:12










  • 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?
    – Sondra
    May 13 '14 at 22:00















up vote
0
down vote













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 $$t=frac{20 log (3)}{log (2)} simeq 31.6993$$






share|cite|improve this answer























  • I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
    – Sondra
    May 12 '14 at 23:49










  • $b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
    – Claude Leibovici
    May 13 '14 at 5:09












  • yeah I got that but why is it to the 20th root? where did you get get the 20?
    – Sondra
    May 13 '14 at 5:10










  • Simplifying the results to get $b$. Logarithms again.
    – Claude Leibovici
    May 13 '14 at 5:12










  • 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?
    – Sondra
    May 13 '14 at 22:00













up vote
0
down vote










up vote
0
down vote









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 $$t=frac{20 log (3)}{log (2)} simeq 31.6993$$






share|cite|improve this 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 $$t=frac{20 log (3)}{log (2)} simeq 31.6993$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 14 '14 at 6:44

























answered May 12 '14 at 7:31









Claude Leibovici

117k1156131




117k1156131












  • I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
    – Sondra
    May 12 '14 at 23:49










  • $b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
    – Claude Leibovici
    May 13 '14 at 5:09












  • yeah I got that but why is it to the 20th root? where did you get get the 20?
    – Sondra
    May 13 '14 at 5:10










  • Simplifying the results to get $b$. Logarithms again.
    – Claude Leibovici
    May 13 '14 at 5:12










  • 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?
    – Sondra
    May 13 '14 at 22:00


















  • I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
    – Sondra
    May 12 '14 at 23:49










  • $b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
    – Claude Leibovici
    May 13 '14 at 5:09












  • yeah I got that but why is it to the 20th root? where did you get get the 20?
    – Sondra
    May 13 '14 at 5:10










  • Simplifying the results to get $b$. Logarithms again.
    – Claude Leibovici
    May 13 '14 at 5:12










  • 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?
    – Sondra
    May 13 '14 at 22:00
















I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
– Sondra
May 12 '14 at 23:49




I get everything except for how you got b to spiders to be that. Where did you 20 in the root come from?
– Sondra
May 12 '14 at 23:49












$b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
– Claude Leibovici
May 13 '14 at 5:09






$b^t=(sqrt[20]{2})^t=3$ Take the logarithms of both sides to get $t=frac{20 log (3)}{log (2)}$.
– Claude Leibovici
May 13 '14 at 5:09














yeah I got that but why is it to the 20th root? where did you get get the 20?
– Sondra
May 13 '14 at 5:10




yeah I got that but why is it to the 20th root? where did you get get the 20?
– Sondra
May 13 '14 at 5:10












Simplifying the results to get $b$. Logarithms again.
– Claude Leibovici
May 13 '14 at 5:12




Simplifying the results to get $b$. Logarithms again.
– Claude Leibovici
May 13 '14 at 5:12












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?
– Sondra
May 13 '14 at 22:00




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?
– Sondra
May 13 '14 at 22:00


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f789788%2fsolving-exponential-function-for-termites-vs-spiders%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa