Logarithms Math 121 Calculus II

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Logarithms

Math 121 Calculus II

D Joyce, Spring 2013

You know all about logarithms already, but one of the best ways to dene and prove properties about them is by means of calculus. We'll do that here. The main reason, however, for going on this excursion is to see how logic is used in formal mathematics. We'll use what we know about calculus to prove statements about logs and exponents. A denition in terms of areas.Consider the area below the standard hyperbolay= 1=x, above thex-axis, and between the vertical linesx= 1 andx=bwherebis a positive number. We'll treat this as a signed area, so that whenb >1 the area is counted positively, and when

0< b <1 we'll count it negatively. In other words, consider the integralZ

b 11x dx.1234

123456

y= 1=x q q qqqqq qqq b Denition 1.Thenatural logarithm, or more simply thelogarithm, of a positive numberb, denoted lnbis dened as lnb=Z b 11x dx:

Properties of the logarithm function.

Theorem 2.ln1 = 0

Proof.Since an integral whose lower limit of integration equals its upper limit of integration is 0, therefore ln1 =Z 1 11x dx= 0: q.e.d. 1 Theorem 3.The function lnxis dierentiable and continuous on its domain (0;1), and its derivative is ddx lnx=1x Proof.By the inverse of the Fundamental Theorem of Calculus, since lnxis dened as an integral, it is dierentiable and its derivative is the integrand 1=x. As every dierentiable function is continuous, therefore lnxis continuous.q.e.d. Theorem 4.The logarithm of a product of two positive numbers is the sum of their loga- rithms, that is, lnxy= lnx+ lny. Proof.We'll use a general principle here that if two functions have the same derivative on an interval and they agree for one particular argument, then they are equal. It's a useful principle that can be used to prove identities like this. Treat the left hand side of the equation as a function ofxleavingyas a constant, thus, f(x) = lnxy. Likewise, let the right hand side of the equation beg(x) = lnx+ lnywhere againyis a constant andxis a variable.

Then, by the chain rule for derivatives,

ddx f(x) =ddx (lnxy) =1xy ddx xy=yxy =1x

We also have

ddx g(x) =ddx (lnx+ lny) =1x + 0 =1x Sincefandghave the same derivatives on the interval (0;1), therefore they dier by a constant. But takingx= 1,f(1) = lnyandg(1) = ln1 + lny= lny, so the constant they dier by is 0, that is to say,f=g.q.e.d. Theorem 5.The logarithm of a quotient of two positive numbers is the dierence of their logarithms, that is, lnx=y= lnxlny. Proof.Although the same kind of proof could be given as in the preceding theorem, we can also derive this from the preceding theorem. Letz=x=yso thatx=yz. Since lnyz= lny+ lnz, therefore lnyzlny= lnz, that is, lnxlny= lnx=y:q.e.d. Theorem 6.The logarithm of the reciprocal of a positive number is the negation of the logarithm of that number, that is, ln1=y=lny. Proof.Using the preceding theorem, ln1=y= ln1lny= 0lny=lny.q.e.d. These theorems can be proved in a more geometric manner using properties of transfor- mations of area. The graph of the hyperbolay= 1=xhas a special property. If you compress the plane vertically by a factor ofc, then expand the plane horizontally by that same factor, then the hyperbola falls on itself. Start with the point (x;1=x), compress vertically to get (x;1=cx), then expand horizontally to get (cx;1=cx), another point on the graph. 2 y= 1=x (c= 3)q aq bq acq bc

The integral

Z b a1x dxdescribes the areaAof the region under the hyperbola above the in- terval [a;b]. When the plane is compressed vertically by a factor ofc, that region is compressed into a region of areaA=c. When it's expanded horizontally, the resulting region expands back to an areaA, but that region is the area under the hyperbola above the interval [ca;cb] which has areaZ cb ca1x dx. Thus Z b a1x dx=Z cb ca1x dx: That translates into the following identity for logarithms lnblna= lncblnca: Settinga= 1,b=x, andc=yyields the identity lnxy= lnx+ lny, while settinga=y, b=x, andc= 1=yyields the identity lnx=y= lnxlny. Theorem 7.Ifnis an integer andxa positive number then lnxn=nlnx. Proof.First, consider the case whenn= 0. Then lnxn= ln1 = 0 = 0lnx=nlnx. Next, consider the case whennis a positive integer. Thennis the sum ofn1's,n=

1 + 1 ++ 1. Therefore,xnis the product ofn x's,xn=xxx, so

lnxn= ln(xxx) = lnx+ lnx++ lnx=nlnx: Finally, consider the case whennis a negative integer. Then lnxn= ln((1=x)n), and sincenis a positive integer, we have by the previous case that ln((1=x)n) =nln(1=x), which equalsnlnx.q.e.d. Theorem 8.Ifnis an integer andxa positive number then lnnpx=1n lnx. Proof.Sincenlnnpx= ln((npx)n) = lnx, divide bynto get the desired identity.q.e.d. Theorem 9.Ifyis an rational number andxa positive number then lnxy=ylnx. 3 Proof.Letybe the rational numberm=nwithnpositive. Then lnxy= lnxm=n= ln(npx)m=mlnnpx=mn lnx=ylnx: q.e.d. Theorem 10.The function lnxis an increasing one-to-one function on its domain (0;1). Proof.Since its derivative 1=xis positive, therefore it's increasing. Every increasing function on an interval is one-to-one.q.e.d. Theorem 11.The graphy= lnxof the function lnxis concave downward. Proof.The second derivative of lnxis1=x2which is negative, therefore its graph is concave downward.q.e.d. Theorem 12.The range of the function lnxincludes all real numbers. Proof.Letbbe any number greater than 1, thenc= lnb >0. Then multiplesncapproach

1asnincreases to1. Butnc=nlnb= lnbn, so the values of the function lnxgrow

arbitrarily large. Also,nc= lnbnapproaches1. Since the function lnxis a continuous function, it takes on all intermediate values as well. Therefore its range includes all real numbers.q.e.d.

Theorem 13.limx!1lnx=1, and lim

x!0+lnx=1: Proof.Those limits hold since the function lnxis an increasing function with domain (0;1) whose range includes all real numbers.q.e.d. We now have enough qualitative information to sketch a graph ofy= lnx.123

123456

y= lnx4

The numbere.

Denition 14.As the function lnxis a one-to-one function with domain (0;1) and range

Logarithms

Math 121 Calculus II

D Joyce, Spring 2013

You know all about logarithms already, but one of the best ways to dene and prove properties about them is by means of calculus. We'll do that here. The main reason, however, for going on this excursion is to see how logic is used in formal mathematics. We'll use what we know about calculus to prove statements about logs and exponents. A denition in terms of areas.Consider the area below the standard hyperbolay= 1=x, above thex-axis, and between the vertical linesx= 1 andx=bwherebis a positive number. We'll treat this as a signed area, so that whenb >1 the area is counted positively, and when

0< b <1 we'll count it negatively. In other words, consider the integralZ

b 11x dx.1234

123456

y= 1=x q q qqqqq qqq b Denition 1.Thenatural logarithm, or more simply thelogarithm, of a positive numberb, denoted lnbis dened as lnb=Z b 11x dx:

Properties of the logarithm function.

Theorem 2.ln1 = 0

Proof.Since an integral whose lower limit of integration equals its upper limit of integration is 0, therefore ln1 =Z 1 11x dx= 0: q.e.d. 1 Theorem 3.The function lnxis dierentiable and continuous on its domain (0;1), and its derivative is ddx lnx=1x Proof.By the inverse of the Fundamental Theorem of Calculus, since lnxis dened as an integral, it is dierentiable and its derivative is the integrand 1=x. As every dierentiable function is continuous, therefore lnxis continuous.q.e.d. Theorem 4.The logarithm of a product of two positive numbers is the sum of their loga- rithms, that is, lnxy= lnx+ lny. Proof.We'll use a general principle here that if two functions have the same derivative on an interval and they agree for one particular argument, then they are equal. It's a useful principle that can be used to prove identities like this. Treat the left hand side of the equation as a function ofxleavingyas a constant, thus, f(x) = lnxy. Likewise, let the right hand side of the equation beg(x) = lnx+ lnywhere againyis a constant andxis a variable.

Then, by the chain rule for derivatives,

ddx f(x) =ddx (lnxy) =1xy ddx xy=yxy =1x

We also have

ddx g(x) =ddx (lnx+ lny) =1x + 0 =1x Sincefandghave the same derivatives on the interval (0;1), therefore they dier by a constant. But takingx= 1,f(1) = lnyandg(1) = ln1 + lny= lny, so the constant they dier by is 0, that is to say,f=g.q.e.d. Theorem 5.The logarithm of a quotient of two positive numbers is the dierence of their logarithms, that is, lnx=y= lnxlny. Proof.Although the same kind of proof could be given as in the preceding theorem, we can also derive this from the preceding theorem. Letz=x=yso thatx=yz. Since lnyz= lny+ lnz, therefore lnyzlny= lnz, that is, lnxlny= lnx=y:q.e.d. Theorem 6.The logarithm of the reciprocal of a positive number is the negation of the logarithm of that number, that is, ln1=y=lny. Proof.Using the preceding theorem, ln1=y= ln1lny= 0lny=lny.q.e.d. These theorems can be proved in a more geometric manner using properties of transfor- mations of area. The graph of the hyperbolay= 1=xhas a special property. If you compress the plane vertically by a factor ofc, then expand the plane horizontally by that same factor, then the hyperbola falls on itself. Start with the point (x;1=x), compress vertically to get (x;1=cx), then expand horizontally to get (cx;1=cx), another point on the graph. 2 y= 1=x (c= 3)q aq bq acq bc

The integral

Z b a1x dxdescribes the areaAof the region under the hyperbola above the in- terval [a;b]. When the plane is compressed vertically by a factor ofc, that region is compressed into a region of areaA=c. When it's expanded horizontally, the resulting region expands back to an areaA, but that region is the area under the hyperbola above the interval [ca;cb] which has areaZ cb ca1x dx. Thus Z b a1x dx=Z cb ca1x dx: That translates into the following identity for logarithms lnblna= lncblnca: Settinga= 1,b=x, andc=yyields the identity lnxy= lnx+ lny, while settinga=y, b=x, andc= 1=yyields the identity lnx=y= lnxlny. Theorem 7.Ifnis an integer andxa positive number then lnxn=nlnx. Proof.First, consider the case whenn= 0. Then lnxn= ln1 = 0 = 0lnx=nlnx. Next, consider the case whennis a positive integer. Thennis the sum ofn1's,n=

1 + 1 ++ 1. Therefore,xnis the product ofn x's,xn=xxx, so

lnxn= ln(xxx) = lnx+ lnx++ lnx=nlnx: Finally, consider the case whennis a negative integer. Then lnxn= ln((1=x)n), and sincenis a positive integer, we have by the previous case that ln((1=x)n) =nln(1=x), which equalsnlnx.q.e.d. Theorem 8.Ifnis an integer andxa positive number then lnnpx=1n lnx. Proof.Sincenlnnpx= ln((npx)n) = lnx, divide bynto get the desired identity.q.e.d. Theorem 9.Ifyis an rational number andxa positive number then lnxy=ylnx. 3 Proof.Letybe the rational numberm=nwithnpositive. Then lnxy= lnxm=n= ln(npx)m=mlnnpx=mn lnx=ylnx: q.e.d. Theorem 10.The function lnxis an increasing one-to-one function on its domain (0;1). Proof.Since its derivative 1=xis positive, therefore it's increasing. Every increasing function on an interval is one-to-one.q.e.d. Theorem 11.The graphy= lnxof the function lnxis concave downward. Proof.The second derivative of lnxis1=x2which is negative, therefore its graph is concave downward.q.e.d. Theorem 12.The range of the function lnxincludes all real numbers. Proof.Letbbe any number greater than 1, thenc= lnb >0. Then multiplesncapproach

1asnincreases to1. Butnc=nlnb= lnbn, so the values of the function lnxgrow

arbitrarily large. Also,nc= lnbnapproaches1. Since the function lnxis a continuous function, it takes on all intermediate values as well. Therefore its range includes all real numbers.q.e.d.

Theorem 13.limx!1lnx=1, and lim

x!0+lnx=1: Proof.Those limits hold since the function lnxis an increasing function with domain (0;1) whose range includes all real numbers.q.e.d. We now have enough qualitative information to sketch a graph ofy= lnx.123

123456

y= lnx4

The numbere.

Denition 14.As the function lnxis a one-to-one function with domain (0;1) and range