THE \BASIC" FUNCTIONS WE WILL ENCOUNTER IN MATH 1013REVIEW/CONTEXTAs dis ussed in lass, every fun tion, no matter how ompli ated, an be built up from a small numberof basi fun tion steps (steps whi h annot be redu ed any further in terms of even simpler sub-steps, orwhi h we understand so well we don't bother to break them down further in this way), ombined withthe operations of

• addition, subtra tion, multipli ation and/or division (\arithmeti " steps);• omposition (taking the output of a previous step and using it as input to the next step);

• inversion (forming the inverse fun tion of an existing fun tion, if the existing fun tion has an inverse).In this se tion of the notes we will list, and dis uss brie y, all of the basi fun tion steps that youwill meet in the ourse. Before we do so, it is worthwhile expanding a bit on the last entry in the\ ompli ating steps" list above. Students who want to get to the main topi right away, an skip overthe aside whi h follows, and ome ba k to it later.

ASIDE REGARDING THE INTERPRETATION OF THE INVERSES

• The third step in the \ ompli ating steps" list above is not stated expli itly in most textbooks. Thisis probably be ause the inverses of basi fun tions are usually interpreted as separate basi fun tionsthemselves in su h textbooks. However, more ompli ated fun tions an also have inverses, so it isgood to in lude this step expli itly in our list.

• Remember that the inverse of a fun tion y = f(x) (if su h an inverse exists) means a fun tion, denotedf−1, su h that, if y = f(x), then x = f−1(y); f−1(y) does NOT mean 1/f(y).

• As dis ussed previously, the inverse of a fun tion is generally a di�erent fun tion than the originalfun tion, i.e., for f−1, the rule for getting from the \new" input (independent variable) value, y, to the\new" output (dependent variable) value, x, is di�erent than the rule, relevant to f, for getting fromthe \original" input (independent variable) value, x, to the \original" output (dependent variable),

y.For example, the ubing fun tion, y = x3 is su h that x31 6= x3

2 whenever x1 6= x2, and hen e y = f(x) = x3DOES have an inverse. This inverse is alled the ube root fun tion, whi h we write in this ontextas x = f−1(y) = y1/3.

• To a mathemati ian, f and f−1, though di�erent fun tions, ontain exa tly the same information (thelinking of ertain pairs of x and y values, the asso iation being from x → y for f and from y → x for

f−1). As soon as one knows f, one also knows f−1 (provided the inverse a tually exists). Thus one an think of f−1 as a \new" fun tion onstru ted from the \old" fun tion, f. A mathemati ian thustypi ally thinks of f−1 as a non-basi fun tion sin e it is onstru ted from a simpler original fun tion

f by the step of \inversion" and hen e will be in lined NOT to in lude, e.g., the ube root fun tionin the list of \basi " fun tions.

• To most �rst year students, the ubing fun tion and the ube root fun tion seem like very \di�erent"fun tions, and, for this reasons, su h students are usually more omfortable with thinking of su han inverse fun tion as also being one of the \basi fun tions".

• In the dis ussion below of the basi fun tion steps we will meet in this ourse, I will make life easierfor most of you by following the perspe tive students usually prefer, and listing the N th root fun tionsand the natural logarithm fun tion, x = ℓn(y) (whi h is a tually the inverse of the exponential fun tion,y = ex) separately as basi fun tions in their own right, even though they are inverses of other basi fun tions in the list.(In the ase of the N th root fun tion, remember that, when N is an even integer, N = 2m, (−x)N = (x)N , sothe inverse fun tion does not exist on the whole of the original domain −∞ < x < ∞. We an, however,make the onventional hoi e of restri ting the original domain to x ≥ 0, so that x = y1/N ≥ 0 be omesuniquely de�ned. With that additional proviso, the N th root exists for all positive integers N.)

• Although we will list some fun tions whi h are really just inverses of other basi fun tions on thelist in the \basi fun tion steps" list below, you should not forget about the in redibly lose relationbetween a fun tion and its inverse, sin e this will be ome important again later when we studydi�erentiation. The se ond example below provides some further things to think about with regardto the inversion step.

• Two examples of a non-basi fun tions whi h, appropriately de�ned, have inverses, serve to makesome of the above dis ussion more on rete.

* Example 1: Consider the fun tion

y = f(x) = x3 + 1 .We an easily solve this equation for x in terms of y, obtaining, as a result, the algebrai expressionfor x = f−1(y). The result is

x = f−1(y) = [y − 1]1/3 .This new fun tion is easy to break down into simpler steps: one starts with y and �rst performsthe arithmeti (subtra tion) step, y → y− 1. Having obtained y− 1, one then, in a se ond step, takesits ube root to get the �nal result. The fun tion is thus a two-step omposite fun tion, with a�rst arithmeti step and a se ond step whi h we an interpret either as a \basi step" (a uberoot), or as an \inversion step" (the inverse of the ube fun tion).* Example 2: We will take sin(x) and cos(x) to be our \basi " trigonometri fun tions in the list below.The tangent fun tion, y = tan(x) an then be thought of as having been obtained from these twobasi fun tions by the arithmeti step of division, sin e y = f(x) = tan(x) = sin(x)/cos(s). If we restri t thedomain of y = tan(x) to −π/2 < x < π/2, then the fa t that tan(x) in reases steadily through this wholeinterval means that, on this interval, the inverse fun tion x = tan−1(y) exists (this is the onventionalde�nition of the inverse tangent). If we in lude the inversion step as one of our \ ompli atingsteps", then one an think of tan−1(y) as being the result of a two-step ompli ating pro ess whi hstarts with two of our basi fun tions, performs an arithmeti ompli ation (division) in the �rststep, and an inversion ompli ation in the se ond step. If we don't in lude \inversion" as oneof our ompli ating steps, then, sin e tan−1(y) is NOT given by sin−1(y)/cos−1(y) or any other su hexpression, we would have to think of tan−1 as a new type of \basi step". We will see, whenwe study di�erentiation, that the perspe tive where we pay attention to the inversion step as atypi al ompli ating step is the more useful perspe tive, allowing us to redu e the number of hard al ulations we a tually have to perform in detail. Until we see su h results, either perspe tiveon this fun tion would be equally good.

A LIST OF THE BASIC FUNCTIONS STEPS WE WILL MEET IN THIS COURSEThe basi fun tions out of whi h any fun tion we meet in this ourse will be \ onstru ted" (using anyof the ompli ating steps above), are

• the positive integer power fun tions, y = xN , and their inverses, the N th root fun tions, x = N√y = y1/N ,with N a positive integer;

• the two basi trig fun tions, sin(x) and cos(x);

• the exponential fun tion y = ex = exp(x), where e is Euler's onstant (a number between 2 and 3 whi hhas a spe ial property we will �nd out about only when study the derivatives of fun tions of thistype) and its inverse, whi h ould be alled exp−1, but, for histori al reasons, is instead alled thenatural logarithm, ℓn, x = ℓn(y).We dis uss ea h of these basi fun tions, and some of the related fun tions that an be onstru ted fromthem by ommon simple ompli ating steps, in the se tions whi h follow.

(1) POWERS, ROOTS, AND RELATED MORE COMPLICATED FUNCTIONS• y = xN , with N a positive integer an a tually be thought of a resulting from an N-fold multipli ation(arithmeti ompli ation) of the simplest fun tion of all, y = x. However, be ause the N th poweroperation is so easy for most of us to understand, we usually treat it as one of our \basi " operations.• If we restri t our attention to x ≥ 0, the inverse fun tions, the N th roots, are de�ned for all su h N.These are written either x = N√

y or x = y1/N , with the latter notation somewhat more ommon thanthe former in most textbooks. The basi meaning of the N th root is as the inverse of the N th power,i.e., x = y1/N MEANS some number x whi h has the property that xN = y.• Many students with Ontario high-s hool ba kgrounds a tually have never had made lear enoughto them the point that the asso iated power relation a tually DEFINES the meaning of the root.To understand a root you always have the option of onverting the problem to the equivalent powerproblem, whi h will typi ally be easier for most of you to understand. For example, to determinewhether or not (27/125)1/3 is equal to 3/5,

– Let y = 27/125 so that (27/125)1/3 = y1/3. We then want to see if this quantity, x, is or is not equal to 3/5.

– But x = y1/3 if and only if y = x3.– Thus the original question is equivalent to seeing whether or not y = 27/125 is equal to (3/5)3, aquestion most students will �nd easier to answer than the original question.

– Finally, we he k the alternate form of the question: (3/5)3 = 33/53 = 27/125, whi h is indeed the y westarted with. Thus 3/5 is indeed the ube root of 27/125 (a statement whi h has the same meaningas the statement that (3/5)3 is equal to 27/125).

• MORE COMPLICATED FUNCTIONS INVOLVING POWERS AND/OR ROOTS* POLYNOMIALS: On e we have the basi positive powers xN , we an multiply them by onstants(an arithmeti ompli ating step) and then, having done that, add (another arithmeti om-pli ating step) su h onstants-times-powers (sometimes alled \monomials") together to form\polynomials". A simple example of this type is y = 3 − 5x + 4x2 − 23x7. A generi polynomial iswritten

y = P (x),= a0 + a1x+ a2x2 + · · ·+ aMxMwhere we are to understand that the a0, · · · , aM are onstants, and only x is variable, and also that,by onvention, the last oeÆ ient, aM , in non-zero. The highest power of x whi h o urs in su ha polynomial is alled the degree of the polynomial. To make sure you understand the generalnotation: for the example, y = 3− 5x+ 4x2 − 23x7, the degree is 7 and

a0 = 3, a1 = −5 a2 = 4, a3 = a4 = a5 = a6 = 0, and a7 = −23 .All polynomials are the result of applying the arithmeti ompli ating steps of multipli ation by onstants, followed by addition, to the basi positive integer power fun tions.

* RATIONAL FUNCTIONS: Having onstru ted polynomials from the basi positive power fun -tions, one an onstru t ratios of polynomials by �rst onstru ting two di�erent polynomials, P (x)and Q(x), and then adding the additional �nal arithmeti ompli ating step of division to formthe even more ompli ated fun tion R(x) = P (x)/Q(x).DEFINITIONS:

∗ A fun tion of the form y = R(x) = P (x)/Q(x), with P (x) and Q(x) both polynomials is alled a\rational" fun tion.

∗ If the degree of P (x) is less than the degree of Q(x), R(x) is alled a \proper rational fun tion".

∗ If the degree of P (x) is greater than or equal to the degree of Q(x), R(x) is alled a \improperrational fun tion".* NEGATIVE INTEGER POWER FUNCTIONS: The meaning of the fun tion y = x−N , with N apositive integer, is, by de�nition, y = 1/xN. This is a tually a two-step omposite fun tion, whi hone an think of as omposed of∗ a �rst step in whi h one forms the positive integer power x → xN = u, and∗ a se ond step, in whi h one takes the output u = xN from the �rst step and divides 1 by it,

u → 1/u = 1/xN .The �rst step is one of our basi steps, while the se ond step, whose algebrai rule is u → 1/u, isan arithmeti ompli ating step.NOTE: though the notation looks formally the same as for the positive integer power fun tion ase, the negative integer power fun tions are in fa t more ompli ated fun tions.The reason it is useful to use the very similar notation is that manipulations whi h are obviouslytrue by simply ounting powers when the powers are all positive integers (xNxM = xN+M and (

xN)M

=

xNM , xM/xN = xM−N for M > N) remain true when the integers are now also allowed to be negative.(You are supposed to have had this all explained to you in high-s hool.)

– POSITIVE FRACTIONAL POWER FUNCTIONS: The fun tion y = xN/M , with N, M both positiveintegers, is also a two-step omposite fun tion, with one step being the N th power fun tion andone step being the M th root fun tion. We assume for simpli ity that N and M have no ommonfa tors. There are two ways of omposing these two steps, both of whi h turn out to give thesame answer (so one is free to use either as a way of thinking about su h fun tions, hoosingwhi hever is most onvenient for the problem at hand).∗ The �rst of the two-step omposite representations of y = xN/M is y =

[

xN]1/M in whi h the �rststep is the power step x → xN = u and the se ond step is the root step u → u1/M =

[

xN]1/M .

∗ The se ond of the two-step omposite representations of y = xN/M is y =[

x1/M]N in whi h the �rststep is the root step x → x1/M = u and the se ond step is the power step u → uN =

[

x1/m]N .

∗ In general, when has two di�erent fun tions, f and g, the order of omposition matters, i.e.,

f ◦ g is not the same as g ◦ f. For example, if f(u) = u2 + 1 and g(v) = sin(v),(f ◦ g)(x) = f(g(x)) = f(sin(x)) = sin2(x) + 1, whereas

(g ◦ f)(x) = g(f(x)) = g(x2 + 1) = sin(x2 + 1) 6= sin2(x) + 1so the two di�erent orders of omposition give di�erent fun tions in this ase. The fa t thatboth orders of omposing the N th power and M th root give the same answer is thus a non-trivialresult, whi h has to be proved. This should have been done for you in high-s hool, assumingyour tea hers were ompetent. (If it wasn't, you an show it by demonstrating that the M thpowers of both expressions give the same result, and hen e that the two expressions must havebeen the same to begin with.)∗ NOTE that, if M is even, the M th root will not be de�ned unless x ≥ 0, so the domain of thepositive fra tional power fun tion in this ase does not in lude negative x.In both of the possible, equivalent orders the positive fra tional power fun tion is obtained by omposing two of our basi steps.

Thus, again, though the notation looks formally the same as for the positive integer powerfun tions, the positive fra tional power fun tions are in fa t more ompli ated fun tions.The reason it is useful to use the very similar looking notation is again be ause, when one doesso, manipulations whi h are obviously true by simply ounting powers when the powers are allpositive integers (xNxM = xN+M and (

xN)M

= xNM , xM/xN = xM−N when M > N) remain true when N and/orM are repla ed by positive fra tions. This is also something you should have had explained toyou in high-s hool.

– NEGATIVE FRACTIONAL POWER FUNCTIONS: The fun tion y = x−N/M, with N, M bothpositive integers, is a three-step omposite fun tion, whi h an be thought of as onstru ted asfollows:

∗ in steps one and two one performs one N th power operation and one M th root operation, om-posing them in either order to get the result u = xN/M , the positive fra tional power result justdis ussed;

∗ in step three, one forms 1/u = 1/xN/M, whi h is denoted x−N/M .The fun tion y = x−N/M thus involves two basi steps (the N th power and M th root steps) and onearithmeti (division) ompli ating step, u → 1/u.On e again, though the notation looks formally the same as for the positive integer power fun -tions, the negative fra tional power fun tions are in fa t (even) more ompli ated fun tions.The reason it is useful to use the very similar looking notation is, also on e more, be ause, whenone does so, manipulations whi h are obviously true by simply ounting powers when the powersare all positive integers (xNxM = xN+M and (

xN)M

= xNM , xM/xN = xM−N when M > N) remain true when

N and/or M are repla ed by positive and/or negative fra tions. This too is something you willhave had explained to you in high-s hool if your tea hers there were ompetent.

FIG. 1:

θbh

aA=2a

B=2bH=2h

Knowing a triangle is a right angle triangle and that one of

hence the SHAPE of the triangle, but NOT its sizefixes the other (non−right) angle andθthe other angles is

(π/2)−θ

HOWEVER, the ratios of sides of the big triangle are the same as the corresponding ratios of sides of the smalltriangle because the extra factor of 2 occurs in both thenumerator and denominator and hence cancels out (i.e.,THE RATIOS ONLY DEPEND ON SHAPE, NOT SIZE)

The shape isdetermined bythe angle θ

(2) THE TRIGONOMETRIC FUNCTIONSThere are six trigonometri (trig) fun tions, sin(θ), cos(θ), tan(θ), csc(θ), sec(θ) and cot(θ). Ea h has as its basi geometri meaning one of the ratios of the di�erent sides of a right-angle triangle, one of whose non-rightangles is θ. (We ould also use our more usual notation, x, for the independent variable, rather than θ.)The reason �xing one of the non-right angles �xes all six possible ratios of di�erent sides of the triangle,and hen e de�nes a fun tion whi h depends on this one angle, is as follows:

FIG. 2:

The geometrical notions "adjacent", "opposite" and "hypoteneuse"

θ

φ

π/2

h

a

b

Hypoteneuse, h, is the side opposite from the right angle

The side a is "opposite" to φ but "adjacent" to θθ but "adjacent" to φThe side b is "opposite" to

φ=(π/2)−θsincethe sum of the 3interior angles isπ

and the relation of internal angles for right angle triangles

• Fixing one of the non-right angles also �xes the other one sin e the sum of the three interior anglesof any triangle is π radians. With the right angle being π/2 radians, if the one non-right angle is θ,the other one then has to be π/2−θ. Thus �xing one non-right angle in the right-angle triangle totally�xes all angles, and hen e totally �xes the shape of the triangle.

• Di�erent triangles with the same shape, but di�erent sizes, di�er by having all sides of the largerversion longer than the orresponding sides of the smaller version by the same ommon fa tor. Aright-angle triangle example in whi h this ommon fa tor is 2 is shown in Figure 1.

• When one takes the ratio of two sides of the bigger of the two triangles with the same shape, the ommon fa tor an els out and one gets the same value as one has for the ratio of the orrespondingsides of the smaller version.

• The expli it geometri de�nitions of the six trig fun tions are given in terms of the geometri allyspe i�ed on epts of \adja ent", \opposite" and \hypoteneuse". A reminder of the meaning of theseterms is given in Figure 2. These de�nitions, whi h should be viewed as spe ifying the most basi meaning of these fun tions, are

sin(θ) = opposite/hypoteneuse

cos(θ) = adjacent/hypoteneuse

tan(θ) = opposite/adjacent

csc(θ) = hypoteneuse/opposite

sec(θ) = hypoteneuse/adjacent

cot(θ) = adjacent/opposite . (1)

• From these de�nitions it is easy to see that, if we take the sine and osine as our \basi " trigfun tions, the other four trig fun tions an be related to them algebrai ally (by the ompli atingstep of division), as follows:tan(θ) = sin(θ)/cos(θ) csc(θ) = 1/sin(θ)

sec(θ) = 1/cos(θ) , cot(θ) = cos(θ)/sin(θ) . (2)

FIG. 3:

��

Another perspective on the basic geometric meaning of

the sine and cosine functions

X

Y

θ

1sin

cos

(x,y)=(cosθ,

θ

θ

)sinθ

An alternate way of thinking about the geometri meaning of our two basi trig fun tions, the sineand osine, is to re ognize that, if one onsiders a triangle whose size is su h that the hypoteneuse is 1,then the opposite side is sin(θ) and the adja ent side is cos(θ). Equivalently, when the hypoteneuse of theright-angle triangle has length 1, the oordinates of the tip of the hypoteneuse, if its tail is at the origin,are (x, y) = (cos(θ), sin(θ)). These two equivalent alternate perspe tives on the meaning of the sine and osinefun tions are illustrated in Figure 3.

This latter perspe tive is the easiest one to use when one wants to generalize the trig fun tions beyondthe values 0 < θ < π/2 (�rst quadrant angles) that are possible for the non-right angles in a right-angletriangle. Expli itly, we DEFINE the two basi fun tions of su h an angle, θ, as follows:• By de�nition, cos(θ) and sin(θ) for non-�rst-quadrant angles, θ, are DEFINED to be su h that the (x, y) oordinates of the tip of a line of length 1, beginning at the origin and pointing in the θ dire tion,are (x, y) = (cos(θ), sin(θ)).

• The other four fun tions for su h θ are de�ned by the algebrai relations given already above, inEqs. (2).

• Some examples of how useful this perspe tive is will be dis ussed brie y in lass. Expli itly* The \CAS rules" used to memorize in whi h quadrants the various trig fun tions are positive ornegative be ome ompletely unne essary.* Remembering the values of the trig fun tions for angles orresponding to dire tions along the oordinate axes be omes something that no longer requires any memorization at all.For more on angular measure, and lots more review of other things you are supposed to already knowabout the trig fun tions, see the link to the long trig fun tion notes �le ontained on the main oursewebpage.

(3) THE EXPONENTIAL FUNCTION AND NATURAL LOGARITHMIf a is any positive number, our review of fra tional powers above shows that aN/M is de�ned for ANYfra tion N/M, positive or negative (remember that we always assume for simpli ity that all ommonfa tors have been an elled, so N and M have no ommon fa tors). a > 0 is required sin e, for a < 0, aN/Mwould be unde�ned whenever M was even.Remembering that any real number has a de imal form, and that any de imal form whi h terminatesis a fra tion, we see that there are fra tions arbitrarily lose to any real number. Thus, the fa t that,when a > 0, a an be raised to any fra tional power means we an also de�ne a fun tion ax, where x annow be ANY real number (either a fra tion or an irrational number). If x is irrational (not a fra tion),we have to de�ne the meaning of ax by taking a limit of aN/M for fra tions N/M that approa h x arbitrarily losely in the limit.With these preliminaries in mind, we have the following de�nitions.DEFINITIONS:

• A fun tion y = ax, with a > 0 a positive onstant and x the independent variable, is alled \theexponential fun tion with base a".• There is a spe ial number, e, lying between 2 and 3, alled Euler's onstant. When the base ofthe exponential fun tion is e, the fun tion y = ex is onventionally alled simply \the exponentialfun tion" (with the hoi e of base a = e then being understood). To understand what is spe ial aboutthe base hoi e a = e requires studying the derivatives of su h fun tions. We will ome ba k to whatit is that makes e spe ial in this regard when we study di�erentiation.

CAUTIONARY NOTE: Students sometimes get onfused about the distin tion between power fun -tions (whether integer or non-integer) and exponential fun tions. The reason is probably be ause bothinvolve one number being raised to the power of another number. However, as fun tions of the givenindependent variable, x, they are very di�erent sin e, for a power fun tion, the independent variable isthe base and the onstant is the power, while for an exponential fun tion, the independent variable isthe power and the onstant is the base. This makes the fun tions behave VERY di�erently as fun tionsof the independent variable x. As an illustration of this fa t, let's onsider the ase that the variable isx and the onstant is a = 3. Then

• the power fun tion is the ubing fun tion, y = x3, while the exponential fun tion is y = 3x;• when we, for example, double x, x → 2x, the power fun tion hanges from x3 → (2x)3 = 23x3 = 8x3, i.e., theold value of the fun tion gets multiplied by the same onstant 8 for all x;• in ontrast, when we double x, x → 2x, the exponential fun tion hanges from 3x → 32x = 3x+x = 3x3x, whi his a non- onstant multiple, 3x, of the original value, 3x, of the fun tion before x was doubled.

• As laimed, the behavior of the two fun tions as fun tions of x is thus radi ally di�erent.

Some properties of exponential fun tions are immediately obvious.

• The positivity of ax for a > 0 and ALL x:* Sin e a > 0, ax is obviously also > 0 for x > 0.

– For x < 0, x = −|x| and a−|x| = 1/a|x|, whi h is also > 0.* a0 = 1 for all a > 0, by de�nition (this de�nition being natural in the ontext of our rules for ombining powers sin e, e.g., 1 = a2/a2 = a2−2 = a0)* Thus ax > 0 for all x, as laimed.(Students who have only pun hed su h things into al uluators and never been told, or never paidattention to, the a tual meaning of the fun tion sometimes think that a positive number raised to anegative power might be negative; we see that this an never happen.)* The \monotoni ity" of ax as a fun tion of x:* If 0 < a < 1, ax DECREASES as x in reases.* If a = 1, sin e 1x = 1 for all x, we have the uninteresting onstant fun tion whose value is 1 for all x.* If a > 1, ax INCREASES as x in reases.Thus, for 0 < a < 1 or a > 1, the fun tion y = f(x) = ax is one-to-one (di�erent x values as input produ edi�erent y values as output) and hen e has an inverse x = f−1(y). This inverse f−1(y) is, for histori alreasons, alled the \logarithm to the base a of y", and written loga(y). Be ause y = ax > 0, the logarithmis not de�ned if y ≤ 0.

• In this ourse, we will mostly use only the exponential fun tion with base e, y = ex, and its logarithm,whi h has the spe ial name \the natural logarithm" and is written in the spe ial notation x = ℓn(y). exif de�ned for all x, but ℓn(y) is de�ned only for y > 0.

• The exponential fun tion, y = ex is sometimes also written y = exp(x). There are two reasons why this an be useful:* The notation allows us to talk about \the fun tion exp" in analogy to the way we talk about\the fun tion sin", \the fun tion os" et ..* If the exponential step o urs as a later step in a omposite fun tion, the argument of theexponential may be ompli ated, e.g.,

f(x) = e

(

sin(x2+1)

x3+2x+1

)

= exp

(

sin(x2 + 1)

x3 + 2x+ 1

)

.In su h a ase, as you an see from the example just given, it an be a lot easier to read andunderstand the se ond form, whi h uses the alternate exp notation.• Sin e exp and ℓn are inverses of one another the usual results for omposing a fun tion and its inverseare valid:* exp (ℓn(y)) = y (obviously true sin e ℓn takes y → x and exp takes that x right ba k to the y it ame from,by the de�nition of the logarithm as the inverse of the exponential fun tion);* ℓn (exp(x)) = x (obviously true sin e exp takes x → y and ℓn then takes that y right ba k to the x it amefrom, again by the basi de�nition of the logarithm).

• The fa t that ℓn is the inverse of exp allows us to easily work out the following properties of the naturallogarithm.

• Remember that, just as we ould onvert questions about the N th root (the inverse of the N th power)to equivalent questions about the N th power, so we an onvert any question about the naturallogarithm (the inverse of the exponential fun tion) to equivalent questions about the exponentialfun tion. Sin e the exponential fun tion simply involves raising a �xed onstant to a variable power,and students usually feel omfortable with power operations, the equivalent form in whi h theexponential fun tion is involved will almost always feel easier to understand.• The key point in obtaining all of the following properties of the natural logarithm is the one-to-onenature of the relation between x and y in y = ex and x = ℓn(y). Thus, if we are given some y and �ndANY x for whi h y = ex, then this x is the only one for whi h this relation is valid and hen e, by thevery de�nition of the natural logarithm, x = ℓn(y).PROPERTIES OF THE NATURAL LOGARITHM• For x, y > 0, ℓn(xy) = ℓn(x) + ℓn(y): To see this, note that* x = eℓn(x) = exp (ℓn(x)) and y = eℓn(y) = exp (ℓn(y)) by the basi de�nition of the natural logarithm.* Similarly, xy = eℓn(xy) by the basi de�nition of the natural logarithm.* The �rst two relations ⇒ xy = eℓn(x) eℓn(y) = eℓn(x)+ℓn(y).* Comparing the expressions for xy in the last two points, it follows that ℓn(xy) = ℓn(x) + ℓn(y), as laimed.* The initial requirement that x and y are both > 0 is be ause ℓn(x) is not de�ned if x ≤ 0 and, similarly,

ℓn(y) is not de�ned if y ≤ 0.Note, however, that if BOTH of x, y are < 0, xy > 0 and, in that ase, ℓn(xy) IS de�ned, even thoughthe RHS of the relation is not. Sin e xy = |x| |y| when x < 0 and y < 0, the appropriate generalizationof the relation in this ase is ℓn(xy) = ℓn(|x|) + ℓn(|y|).

• For x, y > 0, ℓn(

xy

)

= ℓn(x) − ℓn(y): This result is proven in a way entirely analogous to the produ t result.(You should try to onstru t this proof, to make sure you understood the earlier one, if you are notalready familiar with this material be ause of short omings in your high-s hool's math ourses.)As for the produ t ase, the quotient ase an be generalized to the situation where x and y are both< 0, so that x/y > 0 and ℓn(x/y) is still de�ned. The generalization in this ase is ℓn

(

xy

)

= ℓn(|x|) − ℓn(|y|).• For x > 0, and any y, ℓn (xy) = y ℓn(x): This an be seen as follows:* As usual, from the basi de�nition of the natural logarithm, x = eℓn(x) = exp (ℓn(x)) and xy = eℓn(x

y) =

exp [ℓn (xy)].* But xy is also equal to [

eℓnx]y.* By the [ar]

s= ars property of powers, [

eℓnx]y is also equal to ey ℓnx.* Sin e xy = eℓn(x

y) = ey ℓnx, the uniqueness of the exponential representation of xy implies that ℓn (xy) =

y ℓnx, as laimed.* The restri tion to x > 0 is needed be ause (i) ℓn(x) is not de�ned otherwise and (ii) xy annot beformed for general y if x < 0.

• For x > 0, and a > 0, loga(x) = ℓn(x)ℓn(a) : (We don't really need this for the ourse, sin e we will be sti kingto the exponential and natural logarithm, but it is in luded for ompleteness, and to show that it iseasily re onstru ted if needed, so there is no need to memorize it | besides whi h, if you memorize itwithout understanding where it ame from, and then don't use it for a long time, you will probablymisremember it when you go to use it again anyway.) To see where this result omes from:* By the basi de�nitions of the natural logarithm and the logarithm to the base a, x = aloga(x) and

x = eℓn(x).* But, by the basi de�nition of the natural logarithm, a = eℓn(a).* Thus x = aloga(x) =[

eℓn(a)]loga(x)

= e[loga(x) ℓn(a)].* With x = eℓn(x) = e[loga(x) ℓn(a)], the uniqueness of the exponential representation of x implies that

ℓn(x) = loga(x) ℓn(a) and hen e, �nally, that, as laimed,loga(x) =

ℓn(x)

ℓn(a).

• Students who have not been taught (or not paid attention to) the a tual meaning of the naturallogarithm sometimes write down things that an be seen to learly not make sense if one paysattention to the meaning of the natural logarithm. For example, it is easy to see that ℓn(x + y)CANNOT be equal to ℓn(x) + ℓn(y). Here's how:* If it were, e[ℓn(x)+ℓn(y)] would have to be equal to x+ y.* But e[ℓn(x)+ℓn(y)] = eℓn(x) eℓn(y) = xy 6= x+ y.* Thus ℓn(x) + ℓn(y) 6= ℓn(x+ y), as laimed.