The laws of exponents are not mysterious at all. They simply express the way numbers behave in our common experience, and the way they must behave to be consistent with this experience.

First Rule

For example we know that (5^2) (5^4) = (5*5) * (5*5*5*5) = 5^6. This can be generalized as the rule

a^n * a^m = a^(n+m).

This law simply states that when we multiply the product of n factors, each equal to a, by m factors, each also equal to a, we get n+m factors, each equal to a. Expressed just a little more carelessly, when we multiply n a's by m a's we get n+m a's.

This is common sense. It is the way numbers behave. It isn't some law invented to confuse people.

Second Rule

We know that (4^2) ^3 = (4^2)(4^2)(4^2) = (4*4) * (4*4) * (4*4) = 4^6. This is generalized as the rule

(a^n) ^ m = a^ (n*m).

This law says that when we multiply the product of m factors, each of which contains n factors each equal to a, we get n*m factors, each equal to a. A little more loosely, we could say that m factors each with n a's will contain n*m a's.

Third Rule

If 3^2 * x = 3^2, then what could x be but 1?

Since 3^2 * 3^0 = 3^(2+0) = 3^2, what could 3^0 be but 1?

This observation is generalized as the rule

a^0 = 1.

Fourth Rule

We know for sure that a * (1/a) = a/a = 1.

So we know that if a * x = 1, x must be 1/a.

Similarly we know that if a^n * x = 1, then x = 1 / a^n.

So, since 3^2 * 3^-2 = 3^0 = 1, what could 3^-2 be but 1 / 3^2?

These observations generalize to the rule

a^-n = 1 / a^n.

Other rules

Your text contains other rules, and states the rules precisely. You need to know both the reasoning given here for the above rules, and the text statement of the rules. Anything you do with exponents has to follow these rules.

1. Explain each of the above rules in terms of an example using numbers different than those used here. Choose your numbers so that the same number doesn't appear more than once (except as the base) in any given example (e.g., don't use 2^2 * 2^2 = 2^4 as an example; the exponents on the left-hand side match each other and, what is worse, match the base).

Explain why each of the following perversions of the rules of exponents is incorrect, using a simple example with numbers for each:

a^n * b^m = (ab) ^ (n*m)

a^(-n) = - a^n

a^n + a^m = a^(n+m)

a^0 = 0

a^n * a^m = a^(n*m).

Recall the graph of the basic exponential function y = 2^x, shown below:

The basic points of this graph are (0,1) and (1,2). As x becomes large in the negative direction, y approaches zero, so the negative x axis is an asymptote.

The graph of y = A b^x is shown below. The basic points are (0,A) and (1, A*b), and again the negative x axis is an asymptote.

The above graph also depicts another basic property of exponential functions. The points (0,A) and (1, A*b) are separated by a horizontal distance of 1, and the ratio of y values is (A*b) / A. The y values are represented by vertical blue lines, and the ratio (A*b) / A is visualized as the ratio of the lengths of these lines.

To the left, the graph depicts two y values separated by a horizontal distance of 1. The ratio of these y values is identical to the ratio (A*b) / A of the other pair y values. This will be the case for any two graph points as long as their horizontal separation is 1.

The ratio A * b / A is of course just equal to b. It turns out that whenever two graph points are separated by an x distance, or a horizontal distance, of 1, the ratio of y values will be equal to b.

In fact, an exponential function of the form y = A b^x (or of the equivalent form y = A (2^(kx)) ) always has the property that, when two graph points are separated by the some specified distance, the ratio of the y values is the same.

This means that if two graph points are separated by a horizontal distance of, say, 5, and if the y values are in the ratio 9/2, then any two graph points separated by horizontal distance 5 must have y values in the ratio 9/2.

This is an important insight into the behavior of exponential functions.

             y = 1200 (2^(.12 t) )

y = 400 ( 1.07 ) ^ t

y = 250 ( 1 - .12 ) ^ t

y = .04 ( .8 ) ^ t

3. Suppose that y = f(x) = 5 (1.27^x). Find the coordinates of the basic points (0, f(0) ) and (1, f(1) ). The points are separated by a horizontal distance of 1 unit. What is the ratio between the y values?

Now find the coordinates of the points corresponding to x = 3.4 and x = 4.4. These points are also separated by 1 unit. Is the ratio of y values the same as before?

Pick your own two values of x separated by one unit, and verify that the ratio of y values is again the same.

Now pick two x values separated by 2 units. Determine the ratio of y values. Then pick two more x values separated by two units and show that the ratio of the y values is the same.

4. Use the function y = f(x) = 5 (1.27^x) from the preceding problem and write out the expressions for y1 = f(x1) and y2 = f(x1 + 2). Then write out the expression for the ratio y2 / y1. Simplify this expression as much as possible (you may use DERIVE to check your simplification, or even to perform the simplification if you cannot figure out how; but if you don't know how to do it be sure you learn how, because you will be expected to be able to simplify such an expression).

What does your result tell you about how the ratio depends on the x value x1?

Repeat this exercise for y1 = f(x1) and y2 = f(x1 + 1).

5. Recall that the ratio A*b / A of the two basic-point y values of the function y = A b^x is just equal to b, the base of the function.

Find the two basic points of the exponential function y = 3 (2 ^ (.3 x) ). What is the ratio of the two basic-point y values? What therefore would be the y = A b^x form of this function?

What does the value of 2 ^ .3 have to do with this situation?

As we have seen, when the interest rate is r, the growth factor is 1 + r and in any 1-year period the principle will increase by this factor.

If P(n) is the amount after n years, then the amount after 1 more year is P(n+1). Since the growth factor is 1+r, in one year P(n) must grow to (1+r) P(n). It follows that

P(n+1) = (1+r) P(n).

This is the difference equation for a compound-interest situation.

Of course a similar situation also applies to any quantity Q(n) with growth rate r. For any such quantity

Q(n+1) = (1+r) Q(n).

Once we have the difference equation, all we need is an initial amount like P(0) or P(1) [or Q(0) or Q(1)], and we can begin finding the subsequent values in the usual manner.

6. Find P(1), P(2), ..., P(5) for the difference equation P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000.

7. Find Q(1), Q(2), ..., Q(5) for the difference equation Q(n+1) = (1+r) Q(n), with r = -.1 and P(0) = 4.

8. Find Q(n) for n = 1, 2, 3 and 4 for the difference equation Q(n+1) = .85 Q(n), Q(0) = 400. What is the growth rate for this equation?

9. Write a difference equation for the principle if the interest rate is 12%, with an initial principle of $2000. Use this difference equation to find the principle after 1, 2, 3 and 4 years.

10. Write a difference equation for the amount of antibiotic in the body if 8% is removed per hour, and the initial amount is 500 milligrams. Use this equation to find the number of milligrams after 1, 2, 3 and 4 hours.

Observe the temperature of a small hot potato as it approaches room temperature, starting amount 5 minutes after it is removed from an oven or a microwave. Record the temperature vs. the time in minutes since the first reading. Continue making observations until the potato is at least 8 times closer to room temperature than at the first observation. Also record the room temperature.

Graph the temperature vs. time data you have collected, and sketch a smooth curve approximating your data. What is the asymptote for this curve, and what is its significance?

Now modify your data as follows: subtract the room temperature from the temperature of the potato. Call this the 'temperature excess'.

Graph the temperature excess vs. time. Sketch a smooth curve approximating this data. What is the asymptote for this curve, and why should this be the asymptote?

Divide your curve into four segments, each spanning the same horizontal distance.

For each segment, estimate the average slope, or average rate of change, of the temperature excess function over that segment. Then estimate the average temperature excess over the segment. Finally calculate your estimated ratio of average slope to average temperature excess.

How much change is there in the ratio of average slope to average y value?

How long does it take, according to your graph, for the temperature to fall to half its original value? How long does it then take to fall to half of this value?

Choose two points on your temperature excess vs. time curve. Fit the form y = A b^x to these points, but use Texcess instead of y and t instead of x, so your function is Texcess = A b^t. What function do you get?

Plot this function on the same graph as your data. How well does the function fit the data?

How well would you expect a quadratic function to fit the data?

In the preceding section you divided your graph of the data into four segments. Evaluate your function over each of these segments, and from your graph estimate the average value of the function over each segment. What do you get for the ratios of average slope to average temperature excess? How do your results compare with your estimates from the previous section?

How long does it take, according to your graph, for the temperature to fall to half its original value? How long does it then take to fall to half of this value?

What equation would you write to determine the half-life of your temperature function? Simplify this equation as much as possible. Use DERIVE to solve the equation and approximate your answer.

11. Suppose that the temperature excess of an object is given by the function

Texcess(t) = 50 (.97 ^ t).

Sketch a graph of this function and estimate the time required for the temperature excess to fall to 1/8 of its original value.

Divide this time range into 4 equal segments. For each segment determine the ratio of the average rate at which temperature changes to the average temperature excess for the segment. How nearly constant are your ratios?

Estimate the time required for Texcess to fall to half its original value, the additional time required to fall to half this value, and the time required to fall to half this value.

Write the equation you would need to solve to determine the time required for Texcess to fall to half its original value. Simplify this equation as much as possible. Use DERIVE to solve the equation and approximate your answer.

12. If Texcess(t) = 50(.97^t), then if room temperature is  {{ 25 if you used Celsius and 75 if you used Farenheit in your observations }} then what function Temp(t) gives temperature as a function of time?

Your function should be in the generalized form y = A b^x + c. Identify the values of A, b and c.

You should have seen above that the ratio of the rate at which temperature excess changes to the temperature excess is approximately constant, since this ratio is about the same from one segment to the next.

As an example, consider what would happen to the ratios if the temperature excess function was

Texcess(t) = 80 (.98^t).

We would find that:

The time required to fall to 1/8 of the original value is approximately 103 minutes (assuming that t is measured in minutes).

The corresponding four segments are approximately 0 to 26, 26 to 52, 52 to 78 and 78 to 104 minutes. Over the four segments the temperature excess falls from 80 to 47.3 degrees, from 47.3 degrees to 28.0 degrees, from 28.0 degrees to 16.5 degrees, and from 16.5 degrees to 9.8 degrees.

The corresponding rates of change for these four 26-minute intervals are

-32.7 degrees / 26 minutes = -1.26 degrees / minute

-19.3 degrees / 26 minutes = -.74 degrees / minute

-11.5 degrees / 26 minutes = -.44 degrees / minute

-6.7 degrees / 26 minutes = -.26 degrees / minute

The minimum\maximum average temperature excesses for the four intervals are (80+47.3) / 2 = 63.7, (47.3+28.0)/2 = 37.7, (28.0+16.5) / 2 = 22.3, and (16.5 + 9.8) / 2 = 13.2.

If we take the minimum\maximum average temperature excess for each segment as the average temperature excess (this isn't quite accurate because the curve 'sags' in the middle between the endpoints of any segment, so the average is a little lower than the minimum\maximum average temperature), we get the following ratios:

segment 1: rate of change / temp. excess = -1.26 / 63.7 = -.051

segment 2: rate of change / temp. excess = -.74 / 37.7 = -.051

segment 3: rate of change / temp. excess = - .44 / 22.3 = -.051

segment 4: rate of change / temp. excess = -.26 / 13.2 = -.051.

The ratios are identical to two significant figures.

Had the ratios been calculated to 3 significant figures, slight differences would have become apparent. These differences are due to the fact that the intervals over which the ratios are calculated are fairly large, and to the inaccuracy of our estimate of the average temperature excess for each interval.

As you will see in calculus, the ratio of the rate of change dy/dx for an exponential of form y = A b^x to the value y is always the same. Put another way, for such an exponential function, dy / dx is proportional to y.

The temperature excess is proportional to y. It isn't proportional to `sqrt(y) or to y^2 or to y^3, but to y. If it was proportional to `sqrt(y), for example (and dy/dt is in fact proportional to `sqrt(y) for water flowing from a hole in the bottom of a uniform cylinder), then the temperature function would be quadratic and not exponential. If it was proportional to y^2, then the temperature function would be a power function.

For a temperature excess, this has a simple interpretation: the higher the temperature above room temperature, the faster the temperature changes; and the proportionality between the two quantities is direct.

For compound interest the interpretation is equally simple: the more money you have the more interest you earn and the faster your wealth increases, and the proportionality between the rate at which your money changes and the amount you have is direct.

13. For the radioactive decay model Q(t) = 200 (.86 ^ t), divide the range of t values over which the quantity decreases to approximately 1/4 of its original amount into three equal segments. For each segment show that the ratio of the average rate of change of Q to the average value of Q over the interval is in fact nearly constant.

14. If the rate at which an antibiotic is removed from the body is directly proportional to the amount in the body, and if it is observed that the rate of removal is 40 mg/hour when there are 200 milligrams present in the body, then at what rate would antibiotic be removed when there are 70 milligrams present (hint: this is a proportionality problem)?

If measurements indicate that the antibiotic is being removed at the rate of 90 mg/hour, at what rate do we infer the antibiotic is being removed?

Recall that the number of differences required to reduce a sequence to a nonzero constant is the degree of the polynomial required to generate the sequence.

Let's try analysis of differences for the sequence 80, 47.3, 28.0, 16.5, 9.8 of temperature excesses at regular time intervals, as obtained in the preceding section:

Sequence:          80 47.3  28.0 16.5    9.8

First Difference   -33.7    -19.3   -11.5     -6.7

Second Difference:     14.4 7.8   4.8

Third Difference:               -6.8     -3

Clearly the sequence isn't showing any signs of leveling off to a constant. In fact no sequence generated by an exponential function can ever give constant differences (in fact only sequences generated by polynomials can do so).

If, however, we look at the ratios 47.3/80, 28.0/47.3, 16.5/28.0 and 9.8/16.5 of sequence members to their predecessors, we see a very clear pattern:

47.3/80 = .59

28.0/47.3 = .59

16.5/28.0 = .59

9.8 / 16.5 = .59.

This is a property of exponential functions of the form y = A b^x (or y = A (2^(kx) ), or y = A (e^(kx) ). For any such function, if we generate a sequence by evaluating the function at regular intervals, starting from any x value, we obtain a sequence that gives us constant ratios.

This is not a property of sequences generated by exponential functions of the form y = A b^x + c with c not equal to zero (or of either of the other two generalized forms, with c not equal to zero). If, however, we begin by taking the first difference of such a sequence, the ratios of that first-difference sequence will be constant. This is a property of generalized exponential functions.

The problems below will demonstrate the meanings of the above statements.

15. Evaluate the function y = f(x) = 12 (.6 ^ x) for x = 0, 1, 2, 3, 4, and show that the ratios f(1) / f(0), f(2) / f(1), etc. of the sequence members form a constant sequence.

Show that if we add 1 to the values of y, obtaining the function y = g(x) = 12 (.6 ^ x) + 1, then the ratios g(1) / g(0), g(2) / g(1), ... do not form a constant sequence. Show that if the first-difference sequence is first calculated, then the ratios of this sequence are in fact constant and that, furthermore, the constant ratio obtained is the same as for the original f(x) sequence.

16. Return to your original temperature data and your function model for the temperature excess. You were previously asked to give the temperature function that would correspond to your original data. You probably obtained this function by adding the room temperature to your temperature excess function.

Using a reasonable uniform time interval, use your temperature excess function model to generate a sequence of 5 temperature excesses. Show that this sequence has a constant ratio.

Repeat this exercise for the temperature function obtained by adding room temperature to temperature excess, showing this time that the ratio is not constant.