Skip to main content
Logo image

Section 5.4 The logarithm and the exponential

Note: 1 lecture (optional, requires the optional sections Section 3.5, Section 3.6, Section 4.4)
We now have the tools required to properly define the exponential and the logarithm that you know from calculus so well. We start with exponentiation. If n is a positive integer, it is obvious to define
xn:=xβ‹…xβ‹…β‹―β‹…x⏟n times.
It makes sense to define x0:=1. For negative integers, let xβˆ’n:=1/xn. If x>0, define x1/n as the unique positive nth root. Finally, for a rational number n/m (in lowest terms), define
xn/m:=(x1/m)n.
It is not difficult to show we get the same number no matter what representation of n/m we use, so we do not need to use lowest terms.
However, what do we mean by 22? Or xy in general? In particular, what is ex for all x? And how do we solve y=ex for x? This section answers these questions and more.

Subsection 5.4.1 The logarithm

It is convenient to define the logarithm first. Let us show that a unique function with the right properties exists, and only then will we call it the logarithm.

Proof.

To prove existence, we define a candidate and show it satisfies all the properties. Let
L(x):=∫1x1tdt.
Obviously, i holds. Property ii holds via the second form of the fundamental theorem of calculus (Theorem 5.3.3).
To prove property iv, we change variables u=yt to obtain
L(x)=∫1x1tdt=∫yxy1udu=∫1xy1uduβˆ’βˆ«1y1udu=L(xy)βˆ’L(y).
Let us prove iii. Property ii together with the fact that Lβ€²(x)=1/x>0 for x>0, implies that L is strictly increasing and hence one-to-one. Let us show L is onto. As 1/tβ‰₯1/2 when t∈[1,2],
L(2)=∫121tdtβ‰₯1/2.
By induction, iv implies that for n∈N,
L(2n)=L(2)+L(2)+β‹―+L(2)=nL(2).
Given y>0, by the Archimedean property of the real numbers (notice L(2)>0), there is an n∈N such that L(2n)>y. The intermediate value theorem gives an x1∈(1,2n) such that L(x1)=y. Thus (0,∞) is in the image of L. As L is increasing, L(x)>y for all x>2n, and so
limxβ†’βˆžL(x)=∞.
Next 0=L(x/x)=L(x)+L(1/x), and so L(x)=βˆ’L(1/x). Using x=2βˆ’n, we obtain as above that L achieves all negative numbers. And
limxβ†’0L(x)=limxβ†’0βˆ’L(1/x)=limxβ†’βˆžβˆ’L(x)=βˆ’βˆž.
In the limits, note that only x>0 are in the domain of L.
Let us prove v. Fix x>0. As above, iv implies L(xn)=nL(x) for all n∈N. We already found that L(x)=βˆ’L(1/x), so L(xβˆ’n)=βˆ’L(xn)=βˆ’nL(x). Then for m∈N
L(x)=L((x1/m)m)=mL(x1/m).
Putting everything together for n∈Z and m∈N, we have L(xn/m)=nL(x1/m)=(n/m)L(x).
Uniqueness follows using properties i and ii. Via the first form of the fundamental theorem of calculus (Theorem 5.3.1),
L(x)=∫1x1tdt
is the unique function such that L(1)=0 and Lβ€²(x)=1/x.
Having proved that there is a unique function with these properties, we simply define the logarithm or sometimes called the natural logarithm:
ln⁑(x):=L(x).
See Figure 5.6. Mathematicians usually write log⁑(x) instead of ln⁑(x), which is more familiar to calculus students. For all practical purposes, there is only one logarithm: the natural logarithm. See Exercise 5.4.2.

Figure 5.6. Plot of ln⁑(x) together with 1/x, showing the value ln⁑(4).

Subsection 5.4.2 The exponential

Just as with the logarithm we define the exponential via a list of properties.

Proof.

Again, we prove existence of such a function by defining a candidate and proving that it satisfies all the properties. The L=ln defined above is invertible. Let E be the inverse function of L. Property i is immediate.
Property ii follows via the inverse function theorem, in particular via Lemma 4.4.1: L satisfies all the hypotheses of the lemma, and hence
Eβ€²(x)=1Lβ€²(E(x))=E(x).
Let us look at property iii. The function E is strictly increasing since Eβ€²(x)=E(x)>0. As E is the inverse of L, it must also be bijective. To find the limits, we use that E is strictly increasing and onto (0,∞). For every M>0, there is an x0 such that E(x0)=M and E(x)β‰₯M for all xβ‰₯x0. Similarly, for every Ο΅>0, there is an x0 such that E(x0)=Ο΅ and E(x)<Ο΅ for all x<x0. Therefore,
limnβ†’βˆ’βˆžE(x)=0,andlimnβ†’βˆžE(x)=∞.
To prove property iv, we use the corresponding property for the logarithm. Take x,y∈R. As L is bijective, find a and b such that x=L(a) and y=L(b). Then
E(x+y)=E(L(a)+L(b))=E(L(ab))=ab=E(x)E(y).
Property v also follows from the corresponding property of L. Given x∈R, let a be such that x=L(a) and
E(qx)=E(qL(a))=E(L(aq))=aq=E(x)q.
Uniqueness follows from i and ii. Let E and F be two functions satisfying i and ii.
ddx(F(x)E(βˆ’x))=Fβ€²(x)E(βˆ’x)βˆ’Eβ€²(βˆ’x)F(x)=F(x)E(βˆ’x)βˆ’E(βˆ’x)F(x)=0.
Therefore, by Proposition 4.2.6, F(x)E(βˆ’x)=F(0)E(βˆ’0)=1 for all x∈R. Next, 1=E(0)=E(xβˆ’x)=E(x)E(βˆ’x). Then
0=1βˆ’1=F(x)E(βˆ’x)βˆ’E(x)E(βˆ’x)=(F(x)βˆ’E(x))E(βˆ’x).
Finally, E(βˆ’x)β‰ 0
 1 
E is a function into (0,∞) after all. However, E(βˆ’x)β‰ 0 also follows from E(x)E(βˆ’x)=1. Therefore, we can prove uniqueness of E given i and ii, even for functions E:Rβ†’R.
for all x∈R. So F(x)βˆ’E(x)=0 for all x, and we are done.
Having proved E is unique, we define the exponential function (see Figure 5.7) as
exp⁑(x):=E(x).

Figure 5.7. Plot of ex, together with a slope field giving slope y at every point (x,y). The equation ddxex=ex means that y=ex follows these slopes.

If y∈Q and x>0, then
xy=exp⁑(ln⁑(xy))=exp⁑(yln⁑(x)).
We can now make sense of exponentiation xy for arbitrary y∈R; if x>0 and y is irrational, define
xy:=exp⁑(yln⁑(x)).
As exp is continuous, then xy is a continuous function of y. Therefore, we would obtain the same result had we taken a sequence of rational numbers {yn}n=1∞ approaching y and defined xy=limnβ†’βˆžxyn.
Define the number e, called Euler’s number or the base of the natural logarithm, as
e:=exp⁑(1).
Let us justify the notation ex for exp⁑(x):
ex=exp⁑(xln⁑(e))=exp⁑(x).
The properties of the logarithm and the exponential extend to irrational powers. The proof is immediate.

Remark 5.4.4.

There are other equivalent ways to define the exponential and the logarithm. A common way is to define E as the solution to the differential equation Eβ€²(x)=E(x), E(0)=1. See Example 6.3.3, for a sketch of that approach. Yet another approach is to define the exponential function by power series, see Example 6.2.14.

Remark 5.4.5.

We proved the uniqueness of the functions L and E from just the properties L(1)=0, Lβ€²(x)=1/x and the equivalent condition for the exponential Eβ€²(x)=E(x), E(0)=1. Existence also follows from just these properties. Alternatively, uniqueness also follows from the laws of exponents, see the exercises.

Exercises 5.4.3 Exercises

5.4.1.

Given a real number y and b>0, define f:(0,∞)β†’R and g:Rβ†’R as f(x):=xy and g(x):=bx. Show that f and g are differentiable and find their derivative.

5.4.2.

Let b>0, b≠1 be given.
  1. Show that for every y>0, there exists a unique number x such that y=bx. Define the logarithm base b, logb:(0,∞)β†’R, by logb⁑(y):=x.
  2. Show that logb⁑(x)=ln⁑(x)ln⁑(b).
  3. Prove that if c>0, cβ‰ 1, then logb⁑(x)=logc⁑(x)logc⁑(b).
  4. Prove logb⁑(xy)=logb⁑(x)+logb⁑(y), and logb⁑(xy)=ylogb⁑(x).

5.4.4.

Use the geometric sum formula to show (for tβ‰ βˆ’1)
1βˆ’t+t2βˆ’β‹―+(βˆ’1)ntn=11+tβˆ’(βˆ’1)n+1tn+11+t.
Using this fact show
ln⁑(1+x)=βˆ‘n=1∞(βˆ’1)n+1xnn
for all x∈(βˆ’1,1] (note that x=1 is included). Finally, find the limit of the alternating harmonic series
βˆ‘n=1∞(βˆ’1)n+1n=1βˆ’12+13βˆ’14+β‹―

5.4.5.

ex=limnβ†’βˆž(1+xn)n.
Hint: Take the logarithm.
Note: The expression (1+xn)n arises in compound interest calculations. It is the amount of money in a bank account after 1 year if 1 dollar was deposited initially at interest x and the interest was compounded n times during the year. The exponential ex is the result of continuous compounding.

5.4.6.

  1. Prove that for n∈N,
    βˆ‘k=2n1k≀ln⁑(n)β‰€βˆ‘k=1nβˆ’11k.
  2. Prove that the limit
    Ξ³:=limnβ†’βˆž((βˆ‘k=1n1k)βˆ’ln⁑(n))
    exists. This constant is known as the Euler–Mascheroni constant
     2 
    Named for the Swiss mathematician Leonhard Paul Euler (1707–1783) and the Italian mathematician Lorenzo Mascheroni (1750–1800).
    . It is not known if this constant is rational or not. It is approximately Ξ³β‰ˆ0.5772.

5.4.8.

Show that ex is convex, in other words, show that if a≀x≀b, then ex≀eabβˆ’xbβˆ’a+ebxβˆ’abβˆ’a.

5.4.10.

Show that E(x)=ex is the unique continuous function such that E(x+y)=E(x)E(y) and E(1)=e. Similarly, prove that L(x)=ln⁑(x) is the unique continuous function defined on positive x such that L(xy)=L(x)+L(y) and L(e)=1.

5.4.11.

(requires Section 4.3)   Since (ex)β€²=ex, it is easy to see that ex is infinitely differentiable (has derivatives of all orders). Define the function f:Rβ†’R.
f(x):={eβˆ’1/xif x>0,0if x≀0.
  1. Prove that for every m∈N,
    limxβ†’0+eβˆ’1/xxm=0.
  2. Prove that f is infinitely differentiable.
  3. Compute the Taylor series for f at the origin, that is,
    βˆ‘k=0∞f(k)(0)k!xk.
    Show that it converges, but show that it does not converge to f(x) for any given x>0.
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf