Exponential Word Problems (page 2 of 3) Sections: Logbased word problems, exponentialbased word problems Exponential word problems almost always work off the growth / decay formula, A = Pe^{rt}, where "A" is the ending amount of whatever you're dealing with (money, bacteria growing in a petri dish, radioactive decay of an element highlighting your Xray), "P" is the beginning amount of that same "whatever", "r" is the growth or decay rate, and "t" is time. The above formula is related to the compoundinterest formula, and represents the case of the interest being compounded "continuously". Note that the variables may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship: A = Pe^{rt }...or... A = Pe^{kt} ...or... Q = Ne^{kt} ...or... Q = Q_{0}e^{kt} ...and so on and so forth. No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the growth or decay constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.
For this exercise, the units on time t will be hours, because the growth is being measured in terms of hours. The beginning amount P is the amount at time t = 0, so, for this problem, P = 100. The ending amount is A = 450 at t = 6. The only variable I don't have a value for is the growth constant k, which also happens to be what I'm looking for. So I'll plug in all the known values, and then solve for the growth constant: A
= Pe^{kt} The growth constant is 0.25/hour. Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growthdecay constant / time"), as I have displayed above. Note that the constant was positive, because it was a growth constant. If I had come up with a negative answer, I would have known to check my work to find my error. Copyright © Elizabeth Stapel 20022011 All Rights Reserved
In this problem, I know that time "t" will be in hours, because they gave me growth in terms of hours. First, I'll convert "a day and a half" to "thirtysix hours", so my units match. I know that P = 100, and I need to find A at t = 36. But what is the growth constant "k"? And why do they tell me what the doubling time is? They gave me the doubling time because I can use this to find the growth constant k. Then, once I have this constant, I can go on to answer the actual question. So this exercise actually has two unknowns, the growth constant k and the ending amount A. I can use the doubling time to find the growth constant, at which point the only remaining value will be the ending amount, which is what they actually asked for. So first I'll find the constant. If the initial population is 100, then, in 6.5 hours, the population will be 200. I'll set this up and solve for k: A = Pe^{kt} At this point, I need to use logs to solve: ln(2)
= 6.5k I could simplify this to a decimal approximation, but I won't, because I don't want to introduce roundoff error if I can avoid it. So, for now, the growth constant will remain this "exact" value. (I might want to check this value quickly in my calculator, to make sure that this growth constant is positive, as it should be. If I have a negative value at this stage, I need to go back and check my work.) Now that I have the growth constant, I can answer the actual question, which was "How many bacteria will there be in thirtysix hours?" This means using 100 for P, 36 for t, and the above expression for k; then I simplify to find A: A = 100e^{36(ln(2)/6.5)} = 4647.75313957... There will be about 4648 bacteria. You can do a rough check of this answer, using the fact that exponential processes involve doubling (or halving) times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours. If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twentyfour hours, 3200 in thirty hours, and 6400 in thirtysix hours. If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twentyone hours, 1600 in twentyeight hours, and 3200 in thirtyfive hours. The answer we got above, 4678 in thirtysix hours, fits nicely between these two estimates. Warning: When doing the above simplification of 100e^{36(ln(2)/6.5)}, try to do the calculations completely within the calculator in order to avoid roundoff error. It is best to work from the inside out, starting with the exponent, then the exponential, and finally the multiplication, like this: Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2. First, figure out how many doublingtimes that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have: 36 ÷ 6.5 = 72/13 Use this as the power on 2: 100 × 2^{(72/13)} = 4647.75314... Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this trick can be a timesaver. And, yes, you'd use a base of 3 if you'd been given a triplingtime, a base of 4 for a quadruplingtime, etc. << Previous Top  1  2  3  Return to Index Next >>



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