OpenStudy (anonymous):

The initial population is 91. The population doubles every 14 years. What is the equation for the population after t number of years?

5 years ago
OpenStudy (saifoo.khan):

Geometric progression

5 years ago
OpenStudy (saifoo.khan):

Use for formula, \[\Large T_n = ar^{n-1}\]

5 years ago
OpenStudy (saifoo.khan):

Where a is initial popl. i.e. 91.

5 years ago
OpenStudy (anonymous):

every 14 years means use \(A_0\times 2^{\frac{t}{14}}\)

5 years ago
OpenStudy (anonymous):

n(t)=n0e^kt

5 years ago
OpenStudy (anonymous):

n0=initial population

5 years ago
OpenStudy (anonymous):

find k then use k to find t

5 years ago
hero (hero):

There's probably more that one approach to this, but I think @cammyabbo has the more useful one.

5 years ago
OpenStudy (anonymous):

91(2)^(t/14)?

5 years ago
OpenStudy (anonymous):

once you have e, you take the ln of both sides to bring down your exponent

5 years ago
OpenStudy (anonymous):

182=91e^k(2)

5 years ago
OpenStudy (anonymous):

becomes 91=e^k(2)

5 years ago
OpenStudy (anonymous):

to hell with the e, too much work

5 years ago
OpenStudy (anonymous):

thats wrong

5 years ago
OpenStudy (anonymous):

you are told the doubling time, use 2 as the base

5 years ago
OpenStudy (anonymous):

2=e^k(2)

5 years ago
OpenStudy (anonymous):

take the ln of both sides

5 years ago
OpenStudy (anonymous):

ln 2=k(2)

5 years ago
OpenStudy (anonymous):

k=ln 2/2 plug back into original equation to find t

5 years ago
hero (hero):

I don't remember you quoting a general formula @satellite73

5 years ago
OpenStudy (anonymous):

You all have royally confused me. lol

5 years ago
OpenStudy (anonymous):

??

5 years ago
OpenStudy (anonymous):

follow my posts

5 years ago
OpenStudy (anonymous):

you are told the doubling time. doubles every 14 hours right? so it look like this at time 0 you have 91, at time 14 you have \(91\times 2\) at time 28 you have \(91\times 4=91\times 2^2\) and in general you have \(91\times 2^x\)

5 years ago
OpenStudy (anonymous):

now since it doubles every 14 year you know that the exponent should look like \(\frac{t}{14}\) so that if t is a multiple of 14 you get just what you want. now you can make t any value you like so general formula is \[91\times 2^{\frac{t}{14}}\]

5 years ago
hero (hero):

That's not general enough. I'm referring to a global general math formula that you can apply to any problem. I know you're the man on here, but there's a reason why general formulas were created.

5 years ago
OpenStudy (anonymous):

there is no reason not to use 2 as a base if you are doubling

5 years ago
hero (hero):

You always seem to neglect posting such general formulas.

5 years ago
OpenStudy (anonymous):

ok suppose i have a population that increases by 20% every 3 hours. and lets say i start with P then instantly i know i can model this as \[P(1.2)^{\frac{t}{3}}\] what more do i need?

5 years ago
OpenStudy (anonymous):

if if the half life is h, i use \[A_0(\frac{1}{2})^{\frac{t}{h}}\] done

5 years ago
OpenStudy (anonymous):

this going back and forth with e is math teacher nonsense inflicted on students for no good reason at least not at this level

5 years ago
hero (hero):

You could have posted that general formula at the beginning of your explanation. It helps other users understand your basis of approach

5 years ago
OpenStudy (anonymous):

is \[A_0(\frac{1}{2})^{\frac{t}{h}}\] general ?

5 years ago
OpenStudy (anonymous):

but this calls for going straight to the answer using your eyes and head, not a formula doubling time is 14 years, initial population is 91, use \[91\times 2^{\frac{t}{14}}\]an explanation might be needed but a general formula is not

5 years ago
hero (hero):

I don't know if it is or not, but it's better than just presenting some equation where it looks like you just randomly posted some numbers from the problem and threw an equation together. As far as using eyes and head, you're losing sight on the fact that not every user is as familiar with such things as you are.

5 years ago
hero (hero):

You should always post with other users in mind.

5 years ago
hero (hero):

That which is so obvious to you may be so confusing for someone else.

5 years ago
OpenStudy (anonymous):

@Aprilmbug what numbers were you given? a) doubles use the number 2 as a base b) time: 14 years c) initial population, 91 formula : \(91\times 2^{\frac{t}{14}}\) what could be more straightforward?

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

randomly posted numbers? i respectfully disagreed. i used only the numbers written above

5 years ago
hero (hero):

I could write 2 x 14 divided by 91 = t but that would be misleading, wouldn't it?

5 years ago
OpenStudy (anonymous):

compare my formula with the one with the log and the e and all that (which is of course correct, although there is a minor mistake, it should be \(2=e^{14k}\) but no matter

5 years ago
OpenStudy (anonymous):

well the math teachers teach us in these ways

5 years ago
OpenStudy (anonymous):

im not a math wiz, i have to follow general rules and formulas, otherwise i mess up

5 years ago
OpenStudy (anonymous):

and what do you see? a bunch of steps to change the base from 2, which it is because you are givne the doubling time, to the base of e using logs for god sake. why?

5 years ago
OpenStudy (anonymous):

that is one way of solving the problem, i hope it helped

5 years ago
OpenStudy (anonymous):

Yeahh I just went to wolfram alpha. but thanks guys

5 years ago
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