Class 11 Notes I. Exponential functions A. Defintion B. Graphs C. Applications D. Changing exponents / bases E. Finding exponential functions II. Special bases A. 2 B. e III. Exponential equations I. Exponential functions A. Definition Today we'll consider a new type of function: the exponetial function. An exponential function is any function of the form f(x) = a . b^x Notice that this is different than the functions we've been working with so far, where the variable was raised to some power. Here the variable is the power to which we raise something else. An example of an exponential function is f(x) = 2^x When x = 1, we get 2. When x = 2, we get 4. When x = 3, we get 8. Of couse we can also take non-integer or negative values for x: x | f(x) --------- -1 | 1/2 -2 | 1/4 1/2 | sqrt(2) 3/2 | 2 sqrt(2) -1/2 | 1/sqrt(2) The quantity b is called the base of the exponential expression, and it is always positive, while the quantity x is the exponent, and it can be positive or negative. B. Graphs [Sketch some graphs of exponential functions] Exponential function graphs have a few interesting properties. First, they are always positive (if a is positive) or always negative (if a is negative). You can immediately see that this must be the case because they can never cross the axis. There is no x for which b^x = 0 unless b = 0. Second, they are always increasing if b > 1, and always decreasing if b < 1. C. Applications Where do exponential functions come up? They do so whenever we have something that increases or decreases by a constant factor. One very common example is compound interest. Suppose we have an account that earns interest at a rate of 10% per year. We want to know how much it is worth after y years, if we start with a principal p. After one year, the account earns 0.1 p in interest, so its total value is p + 0.1 p = 1.1 p In the second year, the account earns interest again, but this time on its value at the beginning of the year, 1.1p. Thus the interest earned is 0.1 . 1.1p and the value at the end of the second year is 1.1p + 0.1 . 1.1p = 1.1^2 p Clearly there is a pattern. After three years the total value is 1.1^3 p, and the value after y years is p . 1.1^y This is an exponential function with a base of 1.1. In a case like this, the factor 1.1 is called the growth factor. Just 0.1 is the growth rate. In general growth factor = 1 + growth rate. The general formula for interest is value = principal x (1 + growth rate)^t, where t is the number of growth periods (years in this example) that pass. Notice that the growth rate could be negative, in which case the value decreases every year. An example of something that behaves like this is the resale value of a car. The value is the highest when the car is new, and as the car gets older its resale value diminishes. Other applications are very common in the sciences, and we'll talk about them in a bit. D. Changing exponents / bases The example we just went through brings up the issue of the units we use in the exponent. Here's what I mean. Suppose we had an account where the interest was compounded monthly instead of annually, meaning that we get interest every month. To make the example concrete, suppose we have a choice of 1% interest per month, or 12% per year. Are these the same? If not, which is better? For our formula for interest, if we get 1% per month, then after m months the account is worth value = principal (1.01)^m If we get 12% per year, then after y years it is worth value = principal (1.12)^y. To decide which is better, we can realize that after y years have passed, 12y months have passed. Thus the value if we select the monthly compounding option is value = principal (1.01)^(12y) We can rewrite this using our rule for combining exponents. Remember that (x^a)^b = x^(ab). In the same way 1.01^(12y) = (1.01^12)^y Using a calculator, we can find that 1.01^12 = 1.127. Thus the value is value = principal . 1.127^y This is clearly better than 1.12^y, so the monthly option is better. This is an example of a general procedure: changing the exponent. The basic idea is that, if we have f(x) = a . b^(cx), where a, b, and c and numbers, we can always rewrite this to f(x) = a . (b^c)^x. Since b^c is just another number this illustrates an interesting point: we can always rewrite any exponential function in the form f(x) = a . b^x, even if we have (cx) instead of x in the exponent. A particular example of this is a negative exponent. For example f(x) = 2^(-x) is the same as f(x) = 2^(-1.x) = (2^(-1))^x = (1/2)^x We can also use this sort of reasoning to change the base. Suppose we have f(x) = 4^x We can write 4 = 2^2, so this is the same as f(x) = (2^2)^x = 2^(2x). E. Finding exponential functions A last trick with exponential functions is to find an exponential function given a pair of points -- just like any two points determine a unique line, any two points that both have y>0 or y<0 determine an exponential function. Let's take an example. Suppose we have the points (2,2) and (4,8). What exponential function goes through those points? To answer this, we need to find the coefficient a and the base b in f(x) = a . b^x We know that f(2) = a . b^2 = 2 f(4) = a . b^4 = 8 To figure out a and b, we can just divide the second equation by the first one: a . b^4 / a . b^2 = 8 / 2 b^2 = 4 b = 2 Notice that we don't need +-2 here because the base b is always positive. The trick in this procedure is that, by dividing, we cancelled out a and got b by itself. Now we can go back and get a: f(2) = a . 2^2 = 2 a . 4 = 2 a = 1/2 So the function is f(x) = (1/2) 2^x This will work for any pair of points (x1, y1) and (x2, y2) in exactly the same way. We can get the general formula from the same procedure: f(x1) = a . b^x1 = y1 f(x2) = a . b^x2 = y2 b^(x2 - x1) = (y2 / y1) b = (y2 / y1)^[1/(x2 - x1)] Then plugging in: f(x1) = a . b^x1 = y1 b^x1 = y1/a b = (y1/a)^(1/x1) This will give us the exponential function connecting any two points. Notice that you can see whey the points must both have positive y or negative y. We wound up with (y2/y1)^[1/(x2-x1)]. If y2 and y1 have oppositive signs, then y2/y1 is negative, and we don't end up with a real number base. II. Special bases A. 2 Two bases are particularly common in many applications, and they have their own nomenclature, so they're worth mentioning individually. One very common base is 2, and this comes up whenever we talk about some quantity have a doubling time or a half life. Let's take an example from physics. Suppose some radioactive element has a half life of 5 years. What this means is that, in 5 years, half of the original sample will have decayed. Based on this, we can write a formula for the amount that will be left after y years: amount = starting amount x (1/2)^(y/5) Note the 5 in the denominator of the exponent. It's there to ensure that, if we put in y = 5 years, we get back that half the original amount remains. This is a general rule for half-lives: amount = starting amount x (1/2)^(t/half life) If we have a doubling time, it's exactly the same thing, except we change the 1/2 to a 2. As an example, if we have a population of bacteria, we might say that the doubling time of the population is 12 hours. In this case we have number of bacteria = starting number x 2^(t/12 hours) B. e The other base that gets is own discussion is e. The number e is an important mathematical constant like pi. Its value is 2.71828... -- like pi, it goes on for infinitely many digits without ever repeating. The number e comes up whenever we have a process that is continuous, like radioactive atoms decaying or bacteria reproducing. These phenomena always behave according to amount = original amount x e^(k t), where k is a number called the continuous growth rate. It can be negative, in which case the quantity is decaying rather than increasing. Of course we can always change from e to a different base, following the rules we just discussed. III. Exponential equations The final topic for today is exponential equations. Such equations come up quite regularly. Here's an example: suppose an investment of \$1000 has a doubling time of 6 years. How long will it take the investment to grow to \$8000? The first step is to write down the rule of doubling times: value = starting value x 2^(t/doubling time) In this case, we have value = \$1000 x 2^(t/6) We want to know when the value is \$8000, so we must solve \$8000 = \$1000 x 2^(t/6) The first step is to divide both sides by \$1000: 8 = 2^(t/6) The next is to rewrite 8 as an exponential expression with the same base as we have on the other side. In this case, that's 2, and 8 = 2^3: 2^3 = 2^(t/6) Clearly, to solve this we must have t/6 = 3 so the solution is t = 18. It will take 18 years for the value of the investment to rise to \$8000. This is the general solution procedure: 1. Get the exponential expression by itself on one side. 2. Rewrite the other side as an exponential expression with the same base. 3. Set the exponents equal and solve. Practice examples: 125 = 5^(2x) 3 . 2^(x/2) = 48 Notice that this procedure only works if it is possible to rewrite the other side as an exponential with the same base, which requires that the equation be very unusual. We'll learn a more general approach next week.