The function f(x) = e ^ x is used most often as the basic form of the exponential function. Since e lies between 2 and 3 (e is approximately 2.71828), e^x lies somewhere between y = 2^x and y = 3^x.  The function y = 2^x is also used as a basic form of the exponential function. 

The general forms of the exponential function include

f(x) = A (2 ^ (kx)) + c,

f(x) = A b^x + c

and

f(x) = A e^(kx) + c.

The two fundamental examples through which we will begin to understand the behavior of exponential functions include compound interest and radioactive decay. We will later generalize exponential functions through the example of how the temperature of a hot or cold object approaches room temperature.

We will see that all these situations are governed by the fact that the rate of change in the quantity we are exploring (i.e., the principle of an investment, the radioactivity of a sample, the temperature of an object relative to room temperature) is always directly proportional to the quantity present.

These properties of exponential functions will be understood in terms of the laws of exponents, especially the following:

a^b * a^c = a^(b+c)

(a^b) ^ c = a ^ (bc).

We will also look at the doubling time for an increasing exponential (the time required for a given increasing exponential function to double is always the same, regardless of the starting time) and the half-life (the time required for a given decreasing exponential function to lose half its value is always the same regardless of the starting time) of an exponential function. These ideas, and the need to solve exponential equations, will later lead us to define and explore logarithmic functions.

Introductory exercise: Suppose you are given $1000 to invest at 10% interest, with interest to be computed at the end of each year based on principle at the beginning of the year.

How much money will you have after each of the first four years?

How much will you have after 100 years?

Complete this exercise, as best you can within 10 minutes, before you proceed.

You probably didn't have any trouble at all determining that after the first year you will have earned 10% of $1000, or $100, and that your principle will therefore be $1100.

There is a strong tendency to then figure that each year will see the addition of another $100, so that your totals will be $1200, $1300, $1400, etc.. This would especially be the case for anyone who had looked down the page and seen these numbers. However, this would not be correct. For example, during the second year your interest would be based on the $1100 present at the beginning of that year, so interest would be $110. This would bring the total after two years to $1100 + $110 = $1210.

The correct results could be computed as follows:

Year 1: start $1000, interest 10% ($1000) = .1* $1000 = $100, end $1000 + $100 = $1100.

Year 2: start $1100, interest 10% ($1100) = .1* $1100 = $110, end $1100 + $110 = $1210.

Year 3: start $1210, interest 10% ($1210) = .1* $1210 = $121, end $1210 + $121 = $1331.

Year 4: start $1331, interest 10% ($1331) = .1* $1331 = $133.1, end $1331 + $133.1 = $1464.1

We see that every year we add .1 of the beginning principle to that principle to get the ending principle.

Simplifying the Calculation Process

Expressing this yearly calculation in symbols, if P stands for the beginning principle, we add .1 P to P. This gives us

new principle = P + .1 P.

Of course we can apply the distributive law of multiplication over addition to obtain

new principle = P * (1 + .1).

It should be clear that P * (1 + .1) = P + .1 P. Of course 1 + .1 = 1.1, so

new principle = 1.1 P.

We see that the new principle is obtained by multiplying the old principle by 1.1. So our calculation scheme could have been simpler. We could have calculated as follows:

Year 1: Beginning principle $1000, ending principle $1000 * 1.1 = $1100.

Year 2: Beginning principle $1100, ending principle $1100 * 1.1 = $1210.

Year 3: Beginning principle $1210, ending principle $1210 * 1.1 = $1331.

Year 4: Beginning principle $1331, ending principle $1331 * 1.1 = $1464.1.

The 1.1 by which we multiply accounts for both our original principle P and our interest .1 P.

We note that the 4th-year ending principle has been obtained by multiplying by the factor 1.1 a total of 4 times. This is the same as multiplying our original $1000 by 1.1 * 1.1 * 1.1 * 1.1. Use your calculator to verify that the result of multiplying $1000 by 1.1 a total of four times is indeed $1464.1.

Of course $1000 * 1.1 * 1.1 * 1.1 * 1.1 = $1000 * (1.1 ^ 4). You can calculate this quantity directly using your calculator. Once more you should do so in order to validate the result.

This should give you a hint about how to calculate the 100th-year principle. See if you can figure out what to do, and then do the calculation, before you read further.

You probably realized that, over a period of 100 years we will add 10% a total of 100 times, and will therefore multiply by 1.1 a total of 100 times. This is equivalent to multiplying by (1.1) ^ 100. So our 100th-year total will be $1000 (1.1 ^ 100). You can find the number using your calculator.

Important Terminology: In this situation the 10%, which is the same as the proportion .1, is called the growth rate. This makes good sense because here money grows at the rate of 10% per year. The factor 1.1 by which we multiply beginning principle to obtain ending principle is called the growth factor. This also makes good sense, because in mathematics a number by which we multiply is called a factor, and this factor dictates how our principle grows.

The word 'rate' here is not used in the same sense as when we found average rates of changes.  Here the 'rate' is a percent or proportion of the principle; the actual rate of change dy / dx depends on how much principle we have.  This double use of the word 'rate' can be confusing, so think about the distinction carefully.

Exercises 1-4

1. Repeat the introductory exercise for a beginning principle of $5000 and an annual interest rate of 5%. That is, calculate the principle at the end of each of the first 4 years, then calculate the principle at the end of 100 years.

2. Repeat the introductory exercise for a beginning principle of $500 and an annual interest rate of 12%. By what number would you multiply the amount at the beginning of the year to get the amount at the end of the year?

3. Give the expression for the 100-th year ending principle for an original principle of P0 and an interest rate of 6%.

4. What are the growth rate and growth factor for each of the following:

$500 is invested at 15% for 20 years

$30,000 is invested at 7% for 30 years

$2000 is invested at 5% for 40 years.

For each situation, give an expression for the principle after t years.

The rate r at which principle grows determines the growth factor. For example, if the rate is r = 6% = .06, then the growth factor is 1 + .06 = 1.06. A rate of 15% would be r = .15, and the growth factor would be 1 + .15 = 1.15. It is clear then that the growth factor is 1 + r.

If interest is computed for t years, we multiply the original principle by the growth factor a total of t times. For example, if interest rate is 6% for 10 years, then we multiply by the growth factor 1.06 a total of 10 times; this is the same as muptiplying by 1.06 ^ 10. A 15% interest rate for 20 years would entail multiplying the original principle by 1.15 ^ 20. If the interest rate is r, then for the t-year amount we would multiply by (1+r) a total of t times; we would therefore multiply the original principle by (1+r) ^ t.

It is therefore easy to see that if the original principle is P0, then the t-year principle must be

t-year principle = P(t) = P0 (1 + r) ^ t.

We therefore see that the principle is a function P of time t.   Using function notation we write

P(t) = P0 (1+r) ^ t.

For example, if $500 is invested at 15%, we have P0 = $500, r = .15, so growth rate is r = 15% = .15 and growth factor is (1+r) = 1.15. Thus

P(t) = P0 (1+r) ^ t = $500 (1.15) ^ t.

After 20 years we would have

P(20) = $500 (1.15) ^ 20.

You should calculate these amounts and sketch a reasonably accurate graph of P(t) vs. t.

As another example, if $30,000 is invested at 7%, we have P0 = $30000, r = .07, so growth factor r = 7% = .07 and growth rate is (1+r) = 1.07.  Then

P(t) = P0 (1+r) ^ t = $30000 (1.07) ^ t.

After 30 years we would have

30-year amount:  P(30) = $30000 (1.07) ^ 30.

Exercises 5-7

5. For a $200 investment at a 10% annual rate, what are the growth rate and the growth factor? What therefore is the function P(t) that gives principle as a function of time?

For this function determine the principle at t = 0, t = 5, t = 10 and t = 20. Sketch an approximate graph of principle vs. time from t = 0 to t = 20.

How long does it take for the original $200 principle to double to $400?

At what approximate value of t does the principle first reach $300? Starting from that time, how long does it take the principle to double?

At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?

6. Determine the doubling time for a $200 investment at a 20% annual rate and compare to the results of #1. How did a doubling of the rate affect the doubling time of the investment?

7. Determine the doubling time for $2000 investment at a 10% annual rate and compare to the results of #1. How did a ten-fold increase in the initial principle affect the doubling time of the investment?

Suppose the interest rate is 30%, and our original principle is $1. We can easily calculate the principle and interest for the first 4 years. You should quickly calculate these results.

We find that the principle takes values $1, $1.3, $1.69, $2.20 and $2.86. The interest amounts are $.3, $.39, $.51and $.66.

These quantities are apparent in the graph shown below. Note that each principle exceeds the preceding principle by 30% (look at the blue lines), so that the amount by which the principle increases keeps increasing as the principle increases. This can be seen by the nonuniform increases in the lengths of the blue lines. It is worth noting that the semi-transparent red lines, which represent the actual interest amounts, also increase by 30% each year.

The approximate corresponding graph of P(t) = $1(1.3 ^ t) vs. t is shown below. This graph is approximate because points are joined by straight lines rather than by a smooth curve.

The expression P(t) = $1 (1.3 ^ t) can be recognized as an instance of the generalized form f(t) = A b^t + c of an exponential function, with A = $1, b = 1.3 and c = 0.

Exercise 8-9

8. On a single set of coordinate axes, sketch principle vs. time for the first four years, using four different functions, each with an initial principle of $1. Let the rate the 10% for the first function, 20% for the second, 30% for the third and 40% for the fourth.

Does the final principle increase by the same amount when the rate increases from 10% to 20% as it does between 20% and 30%, and is the change in final principle between the 30% and 40% rates the same as the other two? If not what kind of progression is there in the final amounts?

Estimate for each rate the time required to double the principle from the initial $1 to $2. As the percent rate increases in increments of 10%, does the doubling time change by a consistent amount?

9. Repeat the preceding exercise for an initial principle of $5.  You can do this very quickly if you think about how to do it efficiently.

Challenge exercise for calculus-bound students:  The graph shown above connects the points (0,1), (1,1.3), (2,1.69), etc. by straight line segments.   Find the midpoint of each of these segments (halfway between the x values, halfway between the y values) and determine how far the midpoint actually deviates from the corresponding point on the graph of y = 1.3 ^ t, using the midpoint value of t.  Then sketch an approximate graph using the function values at 0, .5, 1, 1.5, ..., connecting the points with straight line segments.  Choose three of these segments, find their midpoints, and make the same comparison between the straight-line approximation and the actual function value.  Did cutting the interval in half double the precision of the approximation, or did it do better (or worse) than that?

Preliminary exercise: Using the year-end values of the principle for the preceding example of a $1 initial investment at interest rate 30%, make your best estimate of the precise time at which the principle P(t) = $1 (1.3^t) first reaches $2, thereby doubling its initial value. Then determine how long it takes to double the 1-year principle of $1.3.

Consider the above function P(t) = $1 (1.3 ^ t), which represents the changing principle on a $1 initial investment at an interest rate of 30%.

Recalling that the year-end values of this investment are $1.3, $1.69 and $2.20 at the ends of the first 3 years, we see that the investment reaches double its initial value at some time between t = 2 and t = 3 years, probably closer to t = 3 than to t = 2.

We see also that the approximate graph crosses the y = 2 line between year 2 and year 3, somewhat closer to year 3 (recall that the above graph is only approximate, using straight lines between data points; a smooth curve would clearly lie below this approximation and would therefore give us a somewhat later doubling time).

We might ask how the doubling time might differ if we started from some instant other than t = 0.  For example, at time t = 1 the principle is $1.3.  How long does it require to double to $2.6?

To see how long it takes to double the $1.3 principle, starting at t = 1, we can again consider the subsequent values $1.69, $2.20, $2.86 of the principle at the ends of succeeding years.

We see that between the end of the second succeeding year, when principle was $2.20, and the end of the third, when the principle was $2.86, the $1.3 had doubled to $2.6. The doubling time is again seen to be between 2 and 3 years, somewhat closer to 3 years.

The graph also shows that the $1.3 increases to $2.6, doubling its value, between t = 3 and t = 4, somewhat closer to t = 4; since the $1.3 value was reached at t = 1, then doubling time is between 'doublingTimeShort = 3-1 = 2 and 'doublingTimeLong = 4 - 1 = 3.

We can obtain an equation for the doubling time `doublingTime starting from t = 0.

The doubling time, starting at time t = 0 with principle P0 = $1, is the time 'doublingTime required to reach 2 P0 or $2, double the original amount.

We see then that the principle after time 'doublingTime, which is P('doublingTime), must therefore be 2 P0. So we can write the equation

P('doublingTime) = $2

to describe this condition. Since for this example P('doublingTime) = $1 (1.3 ^ 'doublingTime), we have

$1 (1.3 ^ 'doublingTime) = $2.

If we divide both sides by $1, we obtain

1.3 ^ 'doublingTime = 2.

This equation can be solved for 'doublingTime. However the solution requires logarithms. We will see soon what logarithms are and how they are used to solve such an equation.

We can also find an equation for the doubling time starting from t = 1:

The doubling time, starting at time t = 1 with principle P(1) = $1.3, is the additional time 'doublingTime required to reach 2 P(1) = $2.6.

At that time, 'doublingTime later than the starting time t = 1, t must be 1 + 'doublingTime. So the principle will be P(1 + 'doublingTime) and we can say that

P(1 + 'doublingTime) = 2 P(1)

or

P(1 + 'doublingTime) = $ 2.6.

Since P(1 + 'doublingTime) = $1 ( 1.3 ^ 'doublingTime), we can write

$1 (1.3 ^ (1 +'doublingTime)) = $2.6.

Dividing both sides by $1, we find that

1.3 ^ (1 + 'doublingTime) = 2.6.

This equation can be simplified using the rule a ^ (b + c) = a^b * a^c:

1.3^1 * 1.3^`doublingTime = 2.6.

Dividing through by 1.3, we obtain

1.3 ^ `doublingTime = 2,

the same equation we obtained for the doubling time starting with t = 0.

Once again this equation can be solved for 'doublingTime using logarithms.

We can symbolize the process to see why the doubling time doesn't depend on when we start:

If we calculate the doubling time starting at some general time t = t0, the condition for doubling is

P(t0 + 'doublingTime) = 2 P(t0).

This will be true for any function P(t).

For the specific function P(t) = $1 (1.3 ^ t), we have

$1 (1.3 ^ (t0 + 'doublingTime) ) = 2 ($1 (1.3 ^ t0) ).

Dividing by the $1 we obtain

1.3 ^ (t0 + 'doublingTime) = 2 (1.3 ^ t0).

This equation can be simplified as before:

1.3 ^ t0  *  1.3 ^ `doublingTime = 2 (1.3 ^ t0),

We can then divide both sides by 1.3 ^ t0 to obtain the familiar result

1.3 ^ `doublingTime = 2.

This shows that the doubling time is independent of the starting time.

Exercise 10-11

10. Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 0, for a $2000 investment at 10%. Simplify this equation as much as possible using valid operations on the equation.

Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time.

11. Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. Simplify this equation as much as possible using valid operations on the equation.

Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time.

You must know the laws of exponents as stated in your text.

You must know the meanings of growth rate and growth factor, and you must be able to explain in terms of these quantities how the form P(t) = P0 (1+r) ^ t arises naturally out of our consideration of compound interest.

The number e appears in one of the standard forms f(t) = A e^(kt) + c of the exponential function. In many senses this is the most standard form, the one on which nearly every mathematician will agree. This number is arguably the most important number in mathematics.

We can understand e in terms of compound interest. Imagine that we start with some principle P0, and some hypothetical investment that yields 100% interest. Of course the growth rate is 100% = 1, so the growth factor is (1 + r) = 1 + 1 = 2. After one year we will have principle 2 P0.

Now suppose that instead of taking the entire 100% at the end of the year, we take the interest in two stages, taking 50% for the first half of the year and then applying the other 50% for the second half. In this case we have a growth factor of (1 + .5) = 1.5, which we get to apply twice. After the first application the principle will be 1.5 P0; applying the growth factor once more will give us

1.5 (1.5 P0) =1.5 ^ 2 P0 = 2.25 P0.

This is the principle at the end of the year. Notice that we now have more money than before, 2.25 P0 vs. 2 P0.

We might be encouraged to split the 100% into more than two parts, to attempt to obtain more money. If we split the 100% into four parts, 25% each, then we would have growth factor (1 + .25) = 1.25. We would be able to apply this factor four times. It should be easy to see that this will result in a year-end principle of

(1.25 ^ 4) P0 = 2.44 P0.

We see the following progression:

1 compounding: year-end principle is (1 + 1) P0 = 2 P0

2 compoundings: year-end principle is (1 + 1/2) ^ 2 P0 = 2 P0

4 compoundings: year-end principle is (1 + 1/4) ^ 4 P0 = 2 P0.

The fractions 1/2 and 1/4 indicate the splitting of the 100% interest into 2 and 4 parts. The exponents 2 and 4 indicate that the number of times we apply the interest is equal to the number of parts into which we divide the 100% interest.

If we were to divide the interest into n parts, then our growth rate for each application would be 1/n and the growth factor would be (1 + 1/n). We would therefore achieve year-end interest

(1 + 1/n) ^ n P0.

We can evaluate the quantity (1 + 1/n) ^ n for increasing values of n. Until we do so it is not clear whether the results will just keep getting larger and larger without limit, or whether they will remain bounded by some limit. It isn't even clear whether the results will continue increasing.

As you will see from the exercises, the numbers do continue increasing, but they also remain bounded.

We have already obtained 2, 2.25 and 2.44 for n = 1, 2 and 4. It turns out that as n gets larger and larger, the results approach a number a little greater than 2.7. This number cannot be represented exactly in decimal form, since it would continue indefinitely with no recurring pattern (such a number is called irrational; `pi and `sqrt(2) are other examples of irrational numbers). To 5 decimal places the number is 2.71828. The next four decimal places make it look like the number is repeating: we get 2.7818281828. But the apparent pattern is just a coincidence, and the number continues with no repeating pattern.

The number approached by (1 + 1/n) ^ n is called e:

e is the large-n limit of the expressioni (1 + 1/n) ^ n.

So if we divide the 100% interest into more and more parts, we approach the year-end principle e * P0, a bit more than 2.7 P0.

12. Use your calculator to evaluate (1 + 1/n) ^ n for n = 2, 4, 10, 100, 1000, and 10000. For each value of n, write down the difference between 2.71828 and your result.   Make a reasonable estimate of what the differences would be for n = 100,000 and for n = 1,000,000.

13. As n continues to increase, (1 + 1/n) ^ n continues to approach 2.71828. However, your calculator will eventually begin to malfunction as you attempt to use larger and larger numbers for n. Most calculators will begin giving smaller and smaller results, and will finally give just 1. This is a result of the approximate nature of the calculator's binary approximation to base-10 arithmetic, and to the limits of its precision.

Determine the approximate value of n at which your calculator begins to give you bad answers. Suggestion: use n = 100,000, then 1,000,000, etc. (just add another 0 each time).

14. Use DERIVE to determine the approximate number n required to obtain the value 2.71828.

Plutonium is very dangerous stuff. It's great for making nuclear bombs. Also, the amount required to almost certainly give you lung cancer, should you inhale it, is most appropriately measured in millionths (maybe billionths) of a gram.

Another thing about plutonium: if you have a sample of the stuff, its radioactivity doesn't decrease all that rapidly. In fact, it takes something like 10,000 years before it loses half of its radioactivity.

Suppose that plutonium loses 7% of its radioactivity in a millennium. This actually means that 7% of the plutonium will change into something else in a millennium. We will therefore have only 93% as much plutonium at the end of the millennium as at the beginning.

Before reading further, suppose we have a sample of 1 gram of plutonium: determine the amount of plutonium left after 1 millennium, then after 2, 3, ..., 10 millennia.

We can approach this problem as a growth rate problem. The only difference between this and compound interest is that the amount of stuff we have is decreasing rather than increasing. We take care of this by simply saying that the rate is negative.

In the present example, where we lose 7% of our stuff every millennium, the rate is -7%, or -.07. The resulting growth factor is thus (1 + r) = (1 + -.07) = .93.  Note that this growth factor is less than 1.  If you multiply by a number less that 1, the value decreases, which is what happens to the activity of a plutonium sample.

Thus, according to this model:

After one millenium we have .93 (1 gram) = .93 grams of plutonium.

After two millenia we have .93 ^2 (1 gram) = .865 grams of plutonium.

After three millenia we have .93 ^3 (1 gram) = .804 grams of plutonium.

After four millenia we have .93 ^4 (1 gram).

After five millenia we have .93 ^5 (1 gram).

After ten millenia we have .93 ^10 (1 gram) = .484 grams of plutonium.

The quantity Q(t) of plutoniuim after t years is

Q(t) = 1 gram (1 + r) ^ t = 1 gram (.93 ^ t).

Exercise 15

15. If the amount of plutonium decreases by 7% per millennium, then how much of a 16-gram sample will remain after 10, 20, 30, 40 and 50 millenia?

Sketch a graph of the amount of plutonium vs. the number of millennia, for t = 0 to 50 millennia.

15.5. If we start with a quantity Q0 of plutonium, and if the amount of plutonium decreases by 7% per millennium, then what expression represents the quantity Q(t) of plutonium after t years (after t years, not after t millenia)?

If we start with quantity Q0 of a radioactive substance, and if the amount changes at rate r (per period), then the growth rate r is negative, the growth factor (1 + r) is less than 1, and the amount after t periods is

Q(t) = Q0 (1 + r) ^ t.

This expression is identical in form to the expression P(t) = P0 (1+r) ^ t for the principle in a compound interest problem. The only difference is that for radioactive decay, r is negative and the growth factor (1+r) is less than 1, both corresponding to a decrease over time in the amount of the substance.

As implied above the half-life is the time required for a quantity to decrease to half its original amount.

As another example, consider an antibiotic in the body. It is gradually filtered out by the kidneys, so that the amount in the bloodstream decreases with time. It turns out that the amount filtered out in a given time tends to be directly proportional to the amount present. Thus the amount in the bloodstream is well approximated by an exponential function of time.

Suppose that an individual receives a dose of an antibiotic which results in an initial quantity of 200 milligrams in the bloodstream. Suppose also that 6% of the quantity initially in the bloodstream has been filtered out after 1 hour.

The growth rate associated with this situation is -6%, so r = -.06. The growth factor is therefore (1 + r) = 1 + -.06 = .94, and the amount remaining after t hours is

Q(t) = 200 mg (.94) ^ t.

The half-life is the time `dt required for Q(t) to drop by half. Starting from t = 0, when Q(t) = 200 mg, the half-life will be the time `dt at which Q(t) reaches 100 mg. So we have

Q(`dt) = 200 mg (.94) ^ `dt = 100 mg.

The resulting equation

200 mg (.94) ^ `dt = 100 mg

can be rearranged (dividing by 200 mg) to give

.94 ^ `dt = 1/2.

This equation will later be solved for `dt using logarithms. For now we can solve it by trial and error, using different values of `dt until we find one that makes the expression .94 ^ `dt sufficiently close to 1/2.

We can use a graph of .94 ^ t vs. t to approximate the desired value of `dt.

We see that .94 ^ t has initial value 0 when t = 0, and falls to .5, half of its initial value, in a time interval of a little over 11 units. We thus see that the half-life of this antibiotic is about 11 hours.

Exercises 16-17

16. The quantity of a certain radioactive element decreases by 15% per day. The initial amount present is 30 grams.

What function Q(t) gives the quantity Q as a function of time t?

Sketch a graph of Q(t) vs. t for 0 <= t <= 10 days. Use this graph to estimate the half-life of the element.

Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.

17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour.

What function Q(t) gives the quantity Q of the antibiotic as a function of the time t since 10:00 a.m.?

How much antibiotic will be present at 3:00 p.m.?

Sketch a graph of Q(t) vs. t for 0 <= t <= 10 hours. Use this graph to estimate the half-life of the element.

Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.

Suggested Exercise:  The value of the half-life of plutonium given here might or might not be very accurate.  Find out what the actual half-life of plutonium really is, and determine how long it would take to reduce an initial quantity to 1% of its original activity.

Challenge Exercise for Calculus-bound students:  Suppose that the activity of a sample of plutonium is initially 5 decays per second.  Using the model above, determine the activity after 5, 10, 15, and 20 millenia.  Then find the approximate average number of decays per second from 0-5 millenia, 5-10 millenia, 10-15 millenia and 15-20 millenia.  Use these averages to determine the total approximate number of decays during each of these time intervals, and the total for the entire 20 millenia.   Speculate on how many decays there will be from the initial time to eternity.

It should be clear that the quantity of a radioactive element decreases in such a way as to eventually approach 0. This is clear from the fact that after one half-life, 1/2 of the original quantity is present, while after two half-lives 1/2 of this quantity, or 1/4 of the original quantity, is present; after three half-lives 1/2 of this, or 1/8 of the original quantity, is present; and so on. After the fourth through tenth half-lives the quantities will be 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, and 1/1024 of the original amount.

From the form Q(t) = Q0 (1+r) ^ t, we can see that if r is negative 1+r is less than l, so that higher and higher powers of (1 + r) will get smaller and smaller. For example if r = -.03, so that 1+r = .97, the t = 1, 2 and 3 powers will be .97^1 = .97, .97^2 = .94 and .97^3 = .91. Every time we multiply by a number less than 1, our result decreases. Furthermore the result decreases toward 0: it doesn't level off at something like .4, or .1, or even .00001, but rather continues decreasing to a value as close to 0 as we might choose.

This means that as t gets large, Q(t) will approach 0. This means that the graph will approach the x axis as an asymptote.

Exercises 17-20

17. For the function Q(t) = Q0 (.9 ^ t), if t = 3, then Q(t) = Q0 (.9^3) = .729 Q0.

Find a value of t for which Q(t) lies between .05 Q0 and .1 Q0.

Find values of t for which Q(t) lies within each of the following ranges:

between .005 Q0 and .01 Q0

between .0005 Q0 and .001 Q0.

Sketch a reasonably accurate graph of Q(t) vs. t, with the range of t sufficient to allow Q(t) to fall below .01 Q0.

In terms of this exercise explain why the positive x axis is a horizontal asymptote for this function.

18. For each of the following functions find a value of t such that Q(t) lies between .05 Q0 and .1 Q0:

Q(t) = Q0 (.8 ^ t)

Q(t) = Q0 (.7 ^ t)

Q(t) = Q0 (.6 ^ t)

Q(t) = Q0 (.5 ^ t).

19. For the function Q(t) = Q0 (1.1^ t), where we note that the growth rate is positive, find a value of t such that Q(t) lies between .05 Q0 and .1 Q0 (hint: the value of t will be negative).

Find values of t for which Q(t) lies within each of the following ranges:

between .005 Q0 and .01 Q0

between .0005 Q0 and .001 Q0.

Sketch a reasonably accurate graph of Q(t) vs. t, with the range of t sufficient to allow Q(t) to fall below .01 Q0.

In terms of this exercise explain why the negative x axis is a horizontal asymptote for this function.

20. For each of the following functions find a value of t such that Q(t) lies between .05 Q0 and .1 Q0:

Q(t) = Q0 (.8 ^ t)

Q(t) = Q0 (.7 ^ t)

Q(t) = Q0 (.6 ^ t)

Q(t) = Q0 (.5 ^ t).

Preliminary exercise: evaluate the two exponential functions f(x) = 5 ( 8^x ) and g(x) = 5 ( 2^(3x) ) for x = 1, 2 and 3. Note the comparison between the values of f(x) and the corresponding values of g(x).

f(x) = 5 ( 8^x) is identical to g(x) = 5 ( 2^(3x) )

As you should have conjectured from the preliminary exercise, the functions

f(x) = 5 ( 8^x ) of form y = A b^x with A = 5, b = 8

and

g(x) = 5 ( 2^(3x) ) of form y = A (2^(kx) ) with A = 5, k = 3

are identical.

Why they are equal:  2^(3x) = 8^x;    2^(kx) = (2^k) ^x

It isn't hard to understand why. By the laws of exponents (specifically (a^b)^c = a^(bc)), we see that

2 ^ (3x) = (2^3) ^ x.

Since 2^3 = 8, we therefore have

2^(3x) = 8^x

So we must have

5 ( 8^x ) = 5 ( 2^(3x) )

from which we see that

f(x) = g(x).

The central idea is that 2 ^ (kx) = (2^k) ^ x. This is just a direct expression of the law of exponents quoted above.

So 2 ^ (kx) can be written as b^x, where b = 2^k.

Therefore the function g(x) = 5 ( 2^(kx) ) can be written as f(x) = 5 ( b ^ x), where again b = 2^k.

This shows that the two forms y = A b^x and y = A ( 2^(kx) ) are in fact identical. We just have to let b = 2^k.

It turns out that the form y = A e^(kx) is also identical to the first form: if we let b = e^k, rather than 2^k, we can easily see that this must be the case.

Since the second and third forms are equivalent to the first, they must be equivalent to one another.

Exercises 21-24

21. What value of b would we use to express each of the following functions in the form y = A b^x?

y = 12 ( 2^(-.5x) )

y = .007 ( 2^(.71 x) )

y = -13 ( 2^(3.9 x) )

Write each of these functions in the form y = A b^x.

22. What value of b would we use to express each of the following functions in the form y = A b^x? (Note: You may use e = 2.718 as a reasonable approximation. Or you may use the e^x key on your calculator. You may even use DERIVE: you get the number e by holding down the ALT key and pressing the E key.)

y = 12 ( e^(-.5x) )

y = .007 ( e^(.71 x) )

y = -13 ( e^(3.9 x) )

Write each of these functions in the form y = A b^x.

23. Try to find a good approximation to the value of k for which the function y = 9 ( 2^(kx) ) is the same as y = 9 ( 13 ^ x ). You will have to use trial and error for now. Soon you will learn to get a precise answer using logarithms.

24. Try to find a good approximation to the value of k for which the function y = 9 ( e^(kx) ) is the same as y = 9 ( 13 ^ x ). You will have to use trial and error for now. Soon you will learn to get a precise answer using logarithms.

Preliminary exercise: What two simultaneous equations do we obtain when we substitute the data points (3,9) and (7,2) into the form y = A ( 2^(kx) )?

Example 1: Fitting y = A b^x to data points (0,8) and (7,4)

Suppose that we have the data points (0,8) and (7,4) to fit to the exponential form y = A b^x. We would obtain the two equations

8 = A * b^0 from point (0,8)

4 = A * b^7 from point (7,4).

We can solve these equations easily. In fact, since b^0 = 1, we see that the first equation gives us

8 = A * b^0 = A*1 = A

from which we conclude that A = 8. Then the second equation gives us

4 = 8 * b^7,

from which we obtain

b^7 = .5

so that

b = .5 ^ (1/7).

This value is easily approximated using a calculator. We obtain b = .9057, rounded to four significant figures.

We therefore obtain the function y = 8 (.9057) ^ x. This could, for example, represent an exponential decay model with growth rate -.0943.

Example 2: Fitting y = A b^x to data points (3,8) and (7,4)

If we wish to fit the points (3,8) and (7,4) with the exponential form y = A b^x, we obtain the equations

8 = A b^3

4 = A b^7.

Unlike the previous example, where we had an x = 0 data point, we cannot directly solve the first equation for A. So we have to work just a little harder. However, the solution is still not particularly difficult.

One thing we can do is divide the first equation by the second. This will eliminate the variable A. We obtain

8 / 4 = (A b^3) / (A b^7)

which simplifies to

2 = b^3 / b^7 = b^-4.

This equation is easily solved for b if we simply raise both sides to the -1/4 power:

( b^-4 ) ^ (-1/4) = 2 ^ (-1/4).

The left-hand side of course gives us b, and the right-hand side can be evaluated with a calculator to give us

b = .83 (rounded to two significant figures).

We can substitute this value of b back into either of the original equations to obtain A.

Example 3: Fitting y = A (2^(kx) ) to data points (3,8) and (7,5)

When we fit the form y = A (2^(kx)) to data points (3,8) and (7,4), we obtain the equations

8 = A (2^(3k) )

5 = A (2^(7k) ).

We can once again divide the first by the second to obtain

(8/5) = [A (2^(3k)) ] / [A (2^(7k))]

which simplifies to

1.6 = [2^(3k)] / [2^(7k)].

Using the law of exponents that states a^b / a^c = a^(b-c), we obtain

1.6 = 2 ^ (-4k).

At this stage we cannot solve this equation for k, since k is stuck in an exponent and none of our multiplications, divisions, additions or subtractions can touch it. We will soon see that logarithms permit us to solve this equation very easily.

Exercises 25-28

Obtain and simplify as far as possible, solving where possible, the system of equations corresponding to each of the following situations:

25. A bacteria culture grows exponentially, according to the form y = A b^x, where y is the area covered by the culture and x is time in hours. Find the model corresponding to a culture which originally covers 12 square centimeters and which, after 8 hours, covers 20 square centimeters.

Graph the resulting function.

Repeat for the form y = A (2^(kx) ).

26. The amount of current flowing through a certain capacitor, as it discharges through the fingers of your left hand, obeys an exponential model of form y = A b^x, where y is current and x is time. The current is 4 microAmps after 2 seconds, and has decreased to 3 microAmps 5 seconds later.

Graph the resulting function.

Repeat for the form y = A (2^(kx) ).