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
It makes sense to define \(x^0 \coloneqq 1\text{.}\) For negative integers, let \(x^{-n} \coloneqq \nicefrac{1}{x^n}\text{.}\) If \(x > 0\text{,}\) define \(x^{1/n}\) as the unique positive \(n\)th root. Finally, for a rational number \(\nicefrac{n}{m}\) (in lowest terms), define
However, what do we mean by \(\sqrt{2}^{\sqrt{2}}\text{?}\) Or \(x^y\) in general? In particular, what is \(e^x\) for all \(x\text{?}\) And how do we solve \(y=e^x\) for \(x\text{?}\) This section answers these questions and more.
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.
Let us prove iii. Property ii together with the fact that \(L'(x) = \nicefrac{1}{x} > 0\) for \(x > 0\text{,}\) implies that \(L\) is strictly increasing and hence one-to-one. Let us show \(L\) is onto. As \(\nicefrac{1}{t} \geq \nicefrac{1}{2}\) when \(t \in [1,2]\text{,}\)
Given \(y > 0\text{,}\) by the Archimedean property of the real numbers (notice \(L(2) > 0\)), there is an \(n \in \N\) such that \(L(2^n) > y\text{.}\) The intermediate value theorem gives an \(x_1 \in (1,2^n)\) such that \(L(x_1) = y\text{.}\) Thus \((0,\infty)\) is in the image of \(L\text{.}\) As \(L\) is increasing, \(L(x) > y\) for all \(x > 2^n\text{,}\) and so
Next \(0 = L(\nicefrac{x}{x}) = L(x) + L(\nicefrac{1}{x})\text{,}\) and so \(L(x) = - L(\nicefrac{1}{x})\text{.}\) Using \(x=2^{-n}\text{,}\) we obtain as above that \(L\) achieves all negative numbers. And
Let us prove v. Fix \(x > 0\text{.}\) As above, iv implies \(L(x^n) = n L(x)\) for all \(n \in \N\text{.}\) We already found that \(L(x) = - L(\nicefrac{1}{x})\text{,}\) so \(L(x^{-n}) = - L(x^n) = -n L(x)\text{.}\) Then for \(m \in \N\)
\begin{equation*}
L(x) = L\Bigl({(x^{1/m})}^m\Bigr) = m L\bigl(x^{1/m}\bigr) .
\end{equation*}
Putting everything together for \(n \in \Z\) and \(m \in \N\text{,}\) we have \(L(x^{n/m}) = n L(x^{1/m}) = (\nicefrac{n}{m}) L(x)\text{.}\)
See Figure 5.6. Mathematicians usually write \(\log(x)\) instead of \(\ln(x)\text{,}\) which is more familiar to calculus students. For all practical purposes, there is only one logarithm: the natural logarithm. See Exercise 5.4.2.
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 \(L\text{.}\) Property i is immediate.
Let us look at property iii. The function \(E\) is strictly increasing since \(E'(x) = E(x) > 0\text{.}\) As \(E\) is the inverse of \(L\text{,}\) it must also be bijective. To find the limits, we use that \(E\) is strictly increasing and onto \((0,\infty)\text{.}\) For every \(M > 0\text{,}\) there is an \(x_0\) such that \(E(x_0) = M\) and \(E(x) \geq M\) for all \(x \geq x_0\text{.}\) Similarly, for every \(\epsilon > 0\text{,}\) there is an \(x_0\) such that \(E(x_0) = \epsilon\) and \(E(x) < \epsilon\) for all \(x < x_0\text{.}\) Therefore,
To prove property iv, we use the corresponding property for the logarithm. Take \(x, y \in \R\text{.}\) As \(L\) is bijective, find \(a\) and \(b\) such that \(x = L(a)\) and \(y = L(b)\text{.}\) Then
\(E\) is a function into \((0,\infty)\) after all. However, \(E(-x) \neq 0\) also follows from \(E(x)E(-x) = 1\text{.}\) Therefore, we can prove uniqueness of \(E\) given i and ii, even for functions \(E \colon \R
\to \R\text{.}\)
for all \(x \in \R\text{.}\) So \(F(x)-E(x) = 0\) for all \(x\text{,}\) and we are done.
Figure5.7.Plot of \(e^x\text{,}\) together with a slope field giving slope \(y\) at every point \((x,y)\text{.}\) The equation \(\frac{d}{dx} e^x = e^x\) means that \(y=e^x\) follows these slopes.
As \(\exp\) is continuous, then \(x^y\) is a continuous function of \(y\text{.}\) Therefore, we would obtain the same result had we taken a sequence of rational numbers \(\{ y_n \}_{n=1}^\infty\) approaching \(y\) and defined \(x^y = \lim_{n\to\infty} x^{y_n}\text{.}\)
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 \(E'(x) = E(x)\text{,}\)\(E(0) = 1\text{.}\) 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.
We proved the uniqueness of the functions \(L\) and \(E\) from just the properties \(L(1)=0\text{,}\)\(L'(x) = \nicefrac{1}{x}\) and the equivalent condition for the exponential \(E'(x) = E(x)\text{,}\)\(E(0) = 1\text{.}\) Existence also follows from just these properties. Alternatively, uniqueness also follows from the laws of exponents, see the exercises.
Given a real number \(y\) and \(b > 0\text{,}\) define \(f \colon (0,\infty) \to \R\) and \(g \colon \R \to \R\) as \(f(x) \coloneqq x^y\) and \(g(x) \coloneqq b^x\text{.}\) Show that \(f\) and \(g\) are differentiable and find their derivative.
Show that for every \(y > 0\text{,}\) there exists a unique number \(x\) such that \(y = b^x\text{.}\) Define the logarithm base \(b\), \(\log_b \colon (0,\infty) \to \R\text{,}\) by \(\log_b(y) \coloneqq x\text{.}\)
Hint: Take the logarithm. Note: The expression \({\left( 1 + \frac{x}{n} \right)}^n\) 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 \(e^x\) is the result of continuous compounding.
Show that \(e^x\) is convex, in other words, show that if \(a \leq x \leq b\text{,}\) then \(e^x \leq e^a \frac{b-x}{b-a} + e^b \frac{x-a}{b-a}\text{.}\)
Show that \(E(x) = e^x\) is the unique continuous function such that \(E(x+y) = E(x)E(y)\) and \(E(1) = e\text{.}\) 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 \(L(e) = 1\text{.}\)
(requires Section 4.3) Since \((e^x)' = e^x\text{,}\) it is easy to see that \(e^x\) is infinitely differentiable (has derivatives of all orders). Define the function \(f \colon \R \to \R\text{.}\)
