Solving the ultimate exponential growth model :D?
86677
Okay, hopefully my quick math is correct....\n\nApril 1st we have 47 rats\nJune 1st we have 900 rats\n\nGrowth model N=No e^k*t  could be re written as:\n\nln(N/No)=k*t\n\nln = natural logarithm\nN = End result of rats\nNo = Beginning amount of rats\nk = growth constant\nt = time\n\nSo, our equation we have \n\nln(900/47) = k*61   (61 is the days from April 1 - June 1)\n\nWhich gives us a growth model of\n\nN=No e^(0.04839*t)\n\nAlrighty...From April 1 - Dec 1 it's approx 240 days\n\nPlugin the numbers  x=amount of rats on Dec 1\n\nln(x/47)=0.04839*240 = 5,197,808.236\n\nWe get approx. 5,197,808 rats on Dec 1st  (alot of rats)\n\nNow you ask when will there be 10000 rats...we plug in our numbers....    t=how many days it'll take\n\nln(10000/47)=0.04839*t = 110.7706 days\n\nSo, approx 111 days from April 1st to reach 10000 rats
739921
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.
105457
Say inflation is constant at 5% per year and you have something worth x.  \n\nOne year later, it is worth x*1.05\nThe next year, it is worth x*1.05*1.05\nThe third year, it is worth x*1.05*1.05*1.05\n\nIn general, for year n, it is worth x * 1.05^n\n\nSo, it is growing exponentially.
473829
The solution set is (-infinity, -7) U (-3/2, infinity)
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
1160088
(1.02)^n = 2, solve for n....ln2 / ln1.02 ~ 35 years
833635
2)\nx and y can increase infinitely as long as the value of x is 1 more than the value of y
131061
After one year the value x will x+x*i=x(1+i) where i the profit and x is the initial value\nAfter 2 year will be x(1+i)^2\nAfter 7 years (2004-1997) will be x(1+i)^7\n\nThe final future value is 1311\nThe initial present value is 309\n(309)(1+i)^7 = 1311\n(1+i)^7 = 1311/309 = 4.2427\n(1+i) = 4.2427^(1/7)\n1+i = 1.22921\ni=0.2292=22.92% annualy\n\nAt 1997- 309*(1+0.229)^0= 309.000000\nAt 1998- 309*(1+0.229)^1= 379.858644\nAt 1999- 309*(1+0.229)^2= 466.9663088\nAt 2000- 309*(1+0.229)^3= 574.0491549\nAt 2001- 309*(1+0.229)^4= 705.6878109\nAt 2002- 309*(1+0.229)^5= 867.5133169\nAt 2003- 309*(1+0.229)^6= 1066.448001\nAt 2004- 309*(1+0.229)^7= 1311.00159
943178
A = A_(0)*exp(k*t) \n\nA represents the amount at a given time t.\n\nAo represents the initial amount of the growing entity.  Note that this is the amount when t = 0.\n\n\nk is a constant that represents the growth rate.  It is POSITIVE when talking in terms of exponential GROWTH.\n\n\nt is the amount of time that has past.  If the information for time is given in dates, you need to convert it to how much time has past since the initial time.  For example, if the model is set up at an initial year of 2000 and you need to find out what the value is in the year 2010, t would be 2010 - 2000 = 10 years.\n\nYou can use this formula to find any of its variables, depending on the information given and what is being asked in a problem.  For example, you may be given the values for Ao and t and you need to find the amount A after the given time.  Or, you may be given the final amount A and the initial amount Ao and you need to find the time t.\n\nSome examples of exponential growth are population growth and financial growth.  The information found, can help predict what a population for a city or colony would be in the future or what the value of your house is in ten years.  Or you can use it to find out how long it would take to get to a certain population or value on your house.
968183
I'm not really sure what you're asking.  The growth rate would depend on the formula itself, so without the formula, there's no way to determine the answer.  You'd need one to get the other.  I imagine that there could be an infinite number of growth rates and in infinite number of growth-rate-formulas.
191015
Since the first 3 answerers didn't really comment on where they got their answers from, I thought a more detailed answer was called for.\n\nThere's a general formula for these types of growth, which also applies to things like radioactive decay rates and interest rates:\n\nP(t) = P(0)*e^(r*t)\n\nP(t) is the population at time t\nP(0) is the population you're starting out with (the population at time 0)\ne is the "natural base" number\nr is the rate of change\nt is the time\n\nIn this case, r is .005 (.5 percent), t is 3 years (878-875), P(0) is 360, and P(t) is our only unknown.  So plug them in:\n\nP(3) = 360*e^(.005*3) = 365.440703...\nWhich rounds to 365.\n\nThe first answerer's method becomes inaccurate after a while; for example, 50 years should be 462, not 450.  The second answerer is better.  The third is just plain wrong, with no explanation of his methods.
480877
i even tried trial and error, there is no solution to this equation. the result of equation can be maximum equla to 174 (98+76), i tried all the possible values by using trial n error , there is no solution to this. SORRY....!!!!
599454
Survival of what exactly?\n\nYou could try an exponential  N=A*exp(-Bt) as used for radioactive decays, with N the nu,ber of survivors.\n\nA logistic function if the numbers of deaths show a normal distribution in time, like for car spare parts.\n\nhttp://en.wikipedia.org/wiki/Logistic_distribution
510279
Jimbo, you gave up too soon......\n\nV = L*W*H = L*W*x in this problem\n\nV = (6-2x)*(4-2x)*x\n\nV  oila!
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
482206
Is exponentially proportionally to the load.
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]