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Exponential Word Problems (page 2 of 3)

Sections: Log-based word problems, exponential-based word problems


Exponential word problems almost always work off the growth / decay formula, A = Pert, 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 X-ray), "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 compound-interest 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 = Pert   ...or...   A = Pekt   ...or...   Q = Nekt   ...or...   Q = Q0ekt

...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.


  • A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant "k" for the bacteria? (Round k to two decimal places.)
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    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 = Pekt 
      450 = 100e6k 
      4.5 = e6k 
      ln(4.5) = 6k 
      ln(4.5)/6 = k = 0.250679566129... 

    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 ("growth-decay 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 2002-2011 All Rights Reserved

  • A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half?
  • 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 "thirty-six 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 = Pekt 
      200 = 100e6.5k 
      2 = e6.5k 

    At this point, I need to use logs to solve:

      ln(2) = 6.5k 
      ln(2)/6.5 = k 

    I could simplify this to a decimal approximation, but I won't, because I don't want to introduce round-off 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 thirty-six hours?" This means using 100 for P, 36 for t, and the above expression for k; then I simplify to find A:

      A = 100e36(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 twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours. If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours. The answer we got above, 4678 in thirty-six hours, fits nicely between these two estimates.

Warning: When doing the above simplification of 100e36(ln(2)/6.5), try to do the calculations completely within the calculator in order to avoid round-off error. It is best to work from the inside out, starting with the exponent, then the exponential, and finally the multiplication, like this:

      calculator steps

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Cite this article as:

Stapel, Elizabeth. "Exponential Word Problems." Purplemath. Available from
    http://www.purplemath.com/modules/expoprob2.htm. Accessed
 

 



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