The Natural Logarithm

Transcription

1 The Natural Logarithm In earlier courses, you may have seen logarithms efine in terms of raising bases to powers. For eample, log 2 8 = 3 because 2 3 = 8. In those terms, the natural logarithm ln = log e shoul be the power to which you raise e to get. (Remember that ln is just shorthan for log e.) Now e = e is represente by an infinite non-repeating ecimal (like π). So, for instance, ln4 is the power to which you raise e = to get 4. How woul you figure that out? You might also woner where e comes from: How o you compute it? An why choose such an ugly number for a logarithm base? In this section, I ll take a ifferent approach to the natural log. I ll efine it using calculus as the area uner a curve. For starters, this allows us to compute its erivative easily. But what oes this have to o with logarithms efine in terms of raising bases to powers? When you stuie logarithms, you learne that they satisfy certain properties: (a) log a y = log a +log a y. (b) log a y = log a log a y. (c) log a p = plog a. () log a = 0. I ll show that the log I efine as an area satisfies those properties. That gives some justification for consiering it to be a logarithm. To begin with, the Power Rule says n = n+ n+ +C for n. The formula oes not apply to. An antierivative F() of woul have to satisfy F() =. But the Funamental Theorem implies that if > 0, then t t =. Thus, t plays the role of F(). t Define the natural log function ln by ln = t for > 0. t

2 By construction, if > 0, ln =, an = ln+c. t t represents the area uner f(t) = t y from to : f(t) = /t ln t ln has many properties you epect a logarithm to have. For eample, ln = t = 0. t You epect the log of a prouct to equal the sum of the logs. If a an b are positive numbers, then ln(ab) = ab a t t = ab t t+ a t t. So In the secon integral, let u = t t, so u =, an t = au. When t = a, u = ; when t = ab, u = b. a a a ab t t+ a a t t = b t t+ u = lna+lnb. u In other wors, In similar fashion, you can verify that ln(ab) = lna+lnb. ln a b = lna lnb an lnp = pln. Thus, there is some justification in calling ln a logarithm, because it has the same properties you epect logs to have. Here are some aitional properties of ln. First, ln = > 0 for > 0. Therefore, the graph of ln is increasing for > 0. Moreover, 2 2 ln = < 0 for > 0. 2 Therefore, the graph of ln is concave own for > 0. 2

3 Net, consier the following picture: f(t) = /t The area uner the curve from to 4 is ln4. It is greater than the sum of the areas of the three rectangles, so ln4 > = 3 2 >. If n is a positive integer, then So if > 4 n, then nln4 > n, or ln4 n > n. ln > ln4 n > n. Since n is an arbitrary positive integer, I can make ln arbitrarily large by making sufficiently large. This proves that lim ln = +. + Here s the graph of ln: y y = ln The ifferentiation formula for ln works together with the other ifferentiation rules in the usual ways. Eample. Compute: (a) ln(2 ++7). (b) ln(sin+3 ). 3

4 (c) [ (ln) 7 +ln( 7 ) ]. 2 () ln. (e) ln(ln(ln+)). (a) (b) (c) () (e) ln(2 ++7) = ln(sin+3 ) = cos+32 sin+ 3. [ (ln) 7 +ln( 7 ) ] = [ (ln) 7 +7ln ] = 7(ln) (ln)(2) ( 2 ) ln = (ln) 2 ( ) ( ln(ln(ln+)) = ln(ln+) = 2ln (ln) 2. )( )( ). ln+ If I say that f() = g() the erivative of f() is g() then g() shoul be efine wherever f() is efine. Therefore, it is not really correct to say without the qualification > 0 that ln =. For is efine for 0, whereas ln is only efine for > 0. It turns out that the correct statement is: ln = for 0. For > 0, this is the same as the ol formula. For < 0, =, so So our new antierivative formula is ln = ln( ) = =. = ln +C. You can omit the absolute value signs if the quantity insie is never negative. For eample, = ln(2 +)+C. Since 2 + is always positive, I can write ln( 2 +) instea of ln

5 Eample. Compute = u u 2 = 2 u u = 2 ln u +C = 2 ln 2 5 +C. ] [ u = 2 5, u = 2, = u 2 Eample. Compute = u u 3 2 = 3 u u = 3 ln u +C = 3 ln C. ] [ u = 3 +2, u = 3 2, = u 3 2 You can use logarithmic ifferentiation to compute erivatives which are ifficult to compute in other ways. Eample. Compute (2 +) 4. Let y = ( 2 +) 4. Taking logs an bringing the power own, I get lny = ln( 2 +) 4 = 4 ln( 2 +). Differentiate both sies, using the Chain Rule on the left an the Prouct Rule (an Chain Rule) on the right: y ( ) 2 y = (4 ) 2 +[ln( 2 +)](4 3 ). + Multiply both sies by y to clear the fraction on the left, then substitute y = ( 2 +) 4 : ( ( ) ) ( ( ) ) 2 2 y = y ( 4 ) 2 +[ln( 2 +)](4 3 ) = ( 2 +) 4 ( 4 ) + 2 +[ln( 2 +)](4 3 ). + I can use calculus to construct the eponential function e. For > 0, I have Since > 0, I have ln = facts: ln =. > 0. This means that ln increases for > 0. Net, I use following (a) An increasing function f is injective: If f(a) = f(b), then a = b. (b) An injective function f : X Y has an inverse function f : Y X. Applying these facts to ln, I fin that it has an inverse function, which I will enote e (or ep()). Thus: 5

6 . e is a function from R to the positive real numbers (because ln is a function from the positive real numbers to R). 2. e ln = for > 0 an lne = for R. It s easy to use the properties of the natural log to erive corresponing properties of e : (a) e +y = e e y. (b) e y = e e y. (c) (e ) p = e p. () e 0 =. To complete the iscussion, I can use ln an e to efine logs an eponentials to other bases. (a) If a > 0 an > 0, efine log a = ln lna. (b) If a > 0, efine a = e lna. With these efinitions, a log a = a (ln)/(lna) = e (ln)/(lna) lna = e ln =. In wors, this says that log a is the power to which you raise a to get the efinition of a log in terms of a base raise to a power that you knew before. Using the efinitions for log a an a, I also get the ifferentiation formulas log a = lna an a = a lna. The point of oing things using calculus is not that we have any new properties. Rather, it allows us to erive properties which were just state an taken for grante in a rigorous way. c 208 by Bruce Ikenaga 6