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I learned calculus for 2 years, but still don't understand the definition of $\ln(x)$

$$\ln(x) = \int_1^x \frac{\mathrm d t}{t}$$

I can't make sense of this definition. How can people find it? Do you have any intuition?

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    $\begingroup$ The derivative of $\ln(x) = \frac{1}{x}$, so by the fundamental theorem of calculus... $\endgroup$
    – mattos
    Mar 6 '18 at 3:53
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    $\begingroup$ The lower limit of integration should be 1. $\endgroup$
    – saulspatz
    Mar 6 '18 at 3:55
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    $\begingroup$ This excellent Numberphile video discusses the need for Euler's number, which is the base of the natural logarithm (i.e. $\ln x = \log_e x$). It should provide some support in defining $\ln$. $\endgroup$
    – CamilB
    Mar 6 '18 at 6:47
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    $\begingroup$ @CamilB With this approach, there is no need for $e$, nor for the notion of base of logarithms. If $\exp$ is the inverse of $\ln$ (it exists because of the properties of differentiable monotonic functions), then $e=\exp(1)$ and one can define $a^x=\exp(a\ln x)$ for $a>0$. The “base $a$” logarithm ensues as inverse function. Of course, some motivation for the abstract definition of $\ln$ is needed anyway. $\endgroup$
    – egreg
    Mar 6 '18 at 10:08
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    $\begingroup$ @ThorstenS.- Who said the integral definition given doesn't define it for $0 < x < 1$? Remember that $$\int_b^a = -\int_a^b$$ $\endgroup$ Mar 6 '18 at 17:35

10 Answers 10

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We want a function that changes multiplication into addition. That is, we want $$f(xy) = f(x) + f(y).\tag 1 $$

Substituting $y=1,$ we get $f(x) = f(x) + f(1),$ so we know that $f(1) = 0.$

Now, let's suppose that $f$ is differentiable. After all, we want to find as nice a function as possible. Let's hold $y$ constant for the moment, and differentiating $(1)$ gives $$yf'(xy) = f'(x) \implies \frac{f'(xy)}{f'(x)}=\frac{1}{y}$$

Now it's not hard to guess that $f'(x) = 1/x$ fills the bill, and together with $f(1)=0,$ the fundamental theorem of calculus gives us the definition.

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    $\begingroup$ There is no need to assume differentiability of $f$ but then the proof is difficult. See math.stackexchange.com/a/2091337/72031 $\endgroup$
    – Paramanand Singh
    Mar 6 '18 at 6:48
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    $\begingroup$ This is how I remember it was taught back when I was still a calculus student. $\endgroup$ Mar 6 '18 at 7:33
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    $\begingroup$ Differentiability need not be assumed, but continuity must be. There are discontinuous functions satisfying (1), if one uses the axiom of choice. $\endgroup$ Mar 7 '18 at 1:43
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    $\begingroup$ @PaulSinclair Granted, but if you were trying to come up with a logarithm for purposes of calculation, you'd look for the smoothest function satisfying the functional equation, wouldn't you? Who wants a nondifferentiable logarithm? It seems to me that solving the functional equation completely is answering a completely different question from the one the OP asked. (Also, I'm answering the question, "How can people find it?" not "How did Napier find it?") $\endgroup$
    – saulspatz
    Mar 7 '18 at 1:50
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    $\begingroup$ I have no problem with your approach. We obviously want a continuous function. I was just expanding on Paramanand Singh's comment that differentiability is not required. (Though obviously you would also want to just assume differentiability with calc 1 students!) $\endgroup$ Mar 7 '18 at 1:56
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Many calculus texts start with the exponential, and the fact that it is equal to its own derivative, and that it has an inverse function if the codomain is taken to be $(0,\infty)$. By the Inverse Function Theorem, if we write $f(x)=e^x$ and $b=e^a$, $$\tag1 (f^{-1})'(b)=\frac1{f'(a)}=\frac1{e^a}=\frac1b. $$ The inverse function of the exponential is usually named $\ln x$, and by $(1)$ we know that $(\ln x)'=1/x$. We also know that $\ln 1=0$, since $e^0=1$. Then $$\tag2\ln x=\int_1^x(\ln t)'\,dt=\int_1^x\frac1t\,dt. $$ The above shows that the natural logarithm should satisfy $(2)$.

Now, it is not easy to come up with the exponential in a constructive way, in particular at an elementary level. So it is easier to start with $(2)$, and construct the exponential as the inverse of $\ln x$.

At a more advanced level, one can start by defining $e^x$ via the Taylor series and then deducing $(2)$ as above. But that wouldn't cut it in a first calculus course.

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The motivation to consider such a function can come from the following observation. We know that $$ \int x^n\,dx=\frac{1}{n+1}x^{n+1}\quad (n\neq-1). $$ Hence we may be interested in an antiderivative of $1/x$. Thus we consider the function (which we name) $$ \log(x)=\int_{1}^x\frac{1}{t}\,dt $$ which has the property that $(\log x)'=\frac{1}{x}$ by the fudamental theorem of calculus. From here we can derive its properties (such as $\log(xy)=\log(x)+\log(y), (x,y>0))$ and realize that the function is indeed the logarithm.

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I must admit this is a non-obvious approach. The idea of logarithm as presented in high school is very simple and it is just considered another way of rewriting the equation $a^x=b$ as $x=\log_{a} b$. The crucial assumption is that $x$ is a rational number and $a>0,a\neq 1$. But such a presentation does not define the symbol $\log_{a} b$ for all $b$. In contrast the log tables systematically list the values of $\log_{10}b$ for all $b$ with $0<b<10$ which can be distinguished using $4$ decimal digits. Clearly the above presentation is very confusing and entirely unsatisfactory if we want to give meaning to symbol $\log_{a} b$ for all real $b>0$.

However the key takeaway from the above presentation is the fact that $$\log_{a} (xy) =\log_{a} x+\log_{a} y$$ whenever each term on both sides of the equation is rational. This gives us some idea to think about functions $f$ for which the equation $$f(xy) =f(x) +f(y) \tag{1}$$ holds generally and then using some assumptions it is possible to show that $f'(x) =k/x$ for some constant $k$ (see this answer). So the functional equation $(1)$ does give us some hints about the derivative of such a function and we can pretty much define the logarithm as an integral (as in your question) and show easily that the functional equation $(1)$ is satisfied.

Another option is to start with the function $f(x) =a^x$ and define it not just for rational $x$ but also for irrational $x$ using limits. Thus if $\{x_n\} $ is a sequence of rational numbers which converges to $x$ then we can define $a^x=\lim_{n\to\infty} a^{x_n}$. This approach is presented in this blog post. In this approach the logarithm arises not as an inverse function but rather springs up as a limit $$\log a=\lim_{h\to 0}\frac{a^h-1}{h}\tag{2}$$ while trying to figure out the derivative of $f(x) =a^x$. Using the limit definition $(2)$ it is easily proved that $\log$ satisfies the functional equation $(1)$ and that $(\log x) '=1/x$ so that $\log x=\int_{1}^{x}t^{-1}\,dt$.

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It all depends on how one defines $\log(x)$ or its inverse $e^x$. If we define $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag1 $$ then because the derivatives converge uniformly on compact sets, we have $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}e^x &=\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1}\\[6pt] &=e^x\tag2 \end{align} $$ Then, the Inverse Function Theorem says that $$ \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}x}\log(x)\right|_{x=a} &=\frac1{\left.\frac{\mathrm{d}}{\mathrm{d}x}e^x\right|_{x=\log(a)}}\\ &=\frac1{\left.\vphantom{\frac{\mathrm{d}}{\mathrm{d}x}}e^x\right|_{x=\log(a)}}\\ &=\frac1a\tag3 \end{align} $$ Equation $(3)$ implies $$ \int_1^x\frac1t\,\mathrm{d}t=\log(x)\tag4 $$

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I am not an expert on the history of mathematics, so take what I say with a grain of salt, but here is my guess as to the historical development of the definition of $\ln(x)$.

Jacob Bernoulli discovered the constant $e$ in studying compound interest. Wikipedia on e

Once you have $e$ defined it is natural to study $e^x$, which also arises in compound interest, and its inverse, $\ln(x)$. So developing the definition of $\ln(x)$ as the inverse of $e^x$ seems the natural way to go, but there are several obstacles which must be overcome if you want to do it rigorously. You would need to define $e^x$ rigorously, show that it has an inverse, show that the inverse must be continuous, etc. So modern calculus texts usually take another route and define $\ln(x)$ as the value of an integral. This gets around most of the problems with rigor, but is usually presented in a way which is generally unmotivated. I'm not surprised that many students find this approach unintuitive.

However, this backwards approach is quite common in mathematics. You start studying an object $A$ and find that it leads to another object, $B$. Then $B$ leads you to $C$. But the path from $A$ to $B$ is long, while the path from $B$ to $C$ is shorter, so you work backwards from $C$ to $B$ and present that as your theorem or definition, carefully erasing all traces of your original path from $A$. This is neat and efficient (in terms of presentation), but it leaves everyone else wondering how the heck you got there.

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  • $\begingroup$ Historically, logarithms appeared way earlier than the exponential. $\endgroup$ Jul 17 '20 at 12:22
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If you have a variable whose range covers many orders of magnitude (like astronomical distances, incomes, historical times) you are interested in a function that "counts the zeros", hence satisfies the functional equation $f(x\,y)=f(x)+f(y)$. Such a function has the property that a percentwise increase of the variable is expressed in a corresponding small additive increment of the function value.

Now it is possible to produce such a function by a simple geometric argument. Consider the graph of the reciprocal function $g(t):={1\over t}$ for $t\geq 1$, and define $f(x)$ as the area under this graph between $t=1$ and $t=x$. Given $x>1$ and $y>1$, stretch the shape defining $f(y)$ horizontally by a factor $x$, so that it now begins at $t=x$ and ends at $t=x\,y$. Then scale it vertically by the factor ${1\over x}$, and you obtain a shape equal to $f(x\, y)-f(x)$.

Many years later you will realize that in fact $f(x)=\int_1^x{1\over t}\> dt\ .$

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I might be stating the obvious here, but:

$$\ln(a) = \int_1^a \frac{dx}{x}$$

means that $\ln(a)$ is defined as the area under the curve $\frac{1}{x}$ between $1$ and $a$.

enter image description here

Note : if $a < 1$, $\ln(a)$ is the opposite of the area under the curve $\frac{1}{x}$ between $a$ and $1$.

From this diagram, we notice that :

  • $\ln(1)$ is $0$ because the blue area will be an empty rectangle.
  • $\ln(x)$ is stricly increasing.
  • $\ln(x) > 0$ when $x > 1$
  • $\ln(x) < 0$ when $x < 1$
  • Somewhere between 2 and 3, the blue area will be as large as one grid square. $\ln(x) = 1$ will have a unique solution, with $2 < x < 3$.
  • $1/x$ becomes infinitely high close to $0$, and infinitely small towards $\infty$. It means $\ln(x)$ is increasing very fast between $0$ and $1$ and very slowly between $1$ and $\infty$.

One possible way to approximate the area is to sum rectangles:

enter image description here

The rectangles are always too large and cover the blue area completely, but the approximation could be made better by taking thinner rectangles.

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The study of the exponential function and the logarithm is motivated by the desire to find a way of modelling a particular kind of growth, where the rate of change of a quantity is proportional to the value of that quantity at any given time. Consider how the population of a colony of bacteria might change over time—perhaps it doubles every hour. Or how your wealth compounds when you put it into a bank. In both cases, we are dealing with exponential growth. Later, when Calculus makes rigorous the notion of 'instantaneous change', we may define an exponential function as a solution to the differential equation $$ \frac{dy}{dx}=ky \, . $$ Solving this equation, we find that $$ kx+C = \int \frac{1}{y} \, dy \, . \tag{*}\label{*} $$ It seems like we have hit a roadblock, since we cannot integrate $1/y$ using the rule $$ \int y^n \, dy = \frac{y^{n+1}}{n+1}+C \, , $$ since that would lead to division by zero. However, the fundamental theorem of calculus tells us that the set of antiderivatives of $1/y$ must be $$ \int_{a}^{y} \frac{1}{t} \, dt \, . $$ This is a set of functions that differ by a constant as we vary the value of $a$. However, since there is already a constant on the LHS of $\eqref{*}$, we can assume that $a=1$ without loss of generality. This yields $$ kx+C = \int_{1}^{y} \frac{1}{t} \, dt \, . $$ Let $f(y)=\int_{1}^{y} \frac{1}{t} \, dt$. This function represents the area of the region bounded by the hyperbola $1/t$, the horizontal axis, and the vertical lines $t=1$ and $t=y$:

Geometric interpretation of f

As the value of $y$ gets larger, the area of this region also gets larger. We can then verify that $f$ is strictly increasing, meaning that $f^{-1}$ exists. Hence, $$ kx+C = f(y) \implies y = f^{-1}(kx+C) \, . $$ Since growth of this kind is called exponential growth, it only seems right to bestow upon the function $f^{-1}$ the new name $\exp$. The statement reads more nicely as $$ y=\exp(kx+C) \, . $$ Of all of the types of exponential growth, the most pleasing is that in which the rate of change of a quantity is equal to the value of that quantity (not just proportional to it). Since the set of solutions to $dy/dx = ky$ is $y=\exp(kx+C)$, the set of solutions to $dy/dx = y$ is $$ y=\exp(x+C) \, . $$ And of these solutions, you will no doubt agree that the simplest and most aesthetic is $$ y=\exp(x) \, , $$ where $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x} \frac{1}{t} \, dt$. For obvious reasons, this function is called the natural exponentional function. There seems to be clear parallels between $\exp$ and functions of the form $g(x)=a^x$. If we return to our examples of exponential modelling—the colony of bacteria that doubles in size every hour—it certainly seems that $2^x$ is an exponential function. After all, the growth of a bacterial colony is proportional to its population at any given point in time. This suggests that $$ 2^x = \exp(kx+C) $$ for some values of $k$ and $C$. (So far, we haven't actually defined what $2^x$ means when $x$ equals $\sqrt{2}$, say, but we'll leave to the side for the moment, and use an 'intuitive' definition of irrational exponents, where $2^{\sqrt{2}} \approx 2^{1.41421}$.) Suspecting that $a^x$ and $\exp(x)$ are somehow linked, we try to recall the essential properties of $a^x$, hoping that $\exp(x)$ shares these properties. The most important of these properties is that $$ a^x \cdot a^y = a^{x+y} \, . $$ And it turns out that $\exp(x)$ also has this property! In other words, $$ \exp(p) \cdot \exp(q) = \exp(p+q) \, . $$ However, proving this takes a little work. Since $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, it seems sensible to use this integral in our proof. Showing that $\exp(p) \cdot \exp(q) = \exp(p+q)$ is actually equivalent to showing that $f(r) + f(s) = f(rs)$, since \begin{align} &f(r) + f(s) = f(rs) \\ \iff &f^{-1}(f(r)+f(s)) = rs \\ \iff &f^{-1}(f(r)+f(s)) = f^{-1}(f(r)) \cdot f^{-1}(f(s)) \\ \iff &\exp(p+q) = \exp(p) \cdot \exp(q) \end{align} (And this should make intuitive sense, too. Notice how $f$ changes multiplication into addition, and so it is only natural that $f^{-1}$ changes addition into multiplication.) There is also a slick way to prove that $f(r)+f(s)=f(rs)$: $$ \int_{1}^{rs}\frac{1}{t} \, dt = \int_{1}^{r}\frac{1}{t} \, dt + \int_{r}^{rs}\frac{1}{t} \, dt $$ Let $z=t/r$, meaning that $dz=\frac{1}{r}dt=\frac{z}{t}dt$. Then we have \begin{align} \int_{1}^{rs}\frac{1}{t} \, dt &= \int_{1}^{r}\frac{1}{t} \, dt + \int_{1}^{s}\frac{1}{z} \, dz \\ f(rs) &= f(r) + f(s) \, . \end{align} Now that we have established that $\exp(p) \cdot \exp(q)=\exp(p+q)$, it seems reasonable to conjecture that $\exp(x)$ is actually equal to $a^x$ for some base $a$. To find that base, we plug in $x=1$, yielding $\exp(1)$, which we will abbreviate as $e$. This means that $$ e^x = \exp(x) \, . $$ This notation, apart from being very convenient, is also entirely valid. To reiterate, $\exp(p) \cdot \exp(q)=\exp(p+q)$, and so $$ e^p \cdot e^q = e^{p+q} $$ as expected. Geometrically, $e$ represents the number $a$ for which $\int_{1}^{a}\frac{1}{t} \, dt = 1$. This property follows directly if one recalls the definition of $\exp$. And since $\log_a(x)$ is defined as the inverse of $a^x$, it is evident that $\log_e(x)$ is the inverse of $e^x$. But we know that the inverse of $\exp(x)$ is $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, and putting these statements together, we find that $$ \log_{e}(x) = \int_{1}^{x}\frac{1}{t} \, dt \, . $$ In fact, this logarithm is so important, so natural, that we often dispense with the base $e$ entirely and write $$ \log(x) = \int_{1}^{x}\frac{1}{t} \, dt \, . $$ Earlier, we mentioned the difficulty of defining $a^x$ when $x$ is an irrational number. The logarithm solves this problem. For rational $x$, we know that $$ a^x = e^{x\log(a)} \, . $$ But the RHS makes sense for all $x$, and so we may define $$ a^x = e^{x\log(a)} \, , $$ for $a>0$. And defining exponents in this way means that $$ a^x \cdot a^y = a^{x+y} \, , $$ still holds even for irrational exponents! The proof follows smoothly if we unravel the definition of $a^x$: $$ a^x = e^{x\log(a)} = \exp(x\log(a)) \, , $$ where $\exp(x)=\log^{-1}(x)$ and $\log(x) = \int_{1}^{x}\frac{1}{t} \, dt$. Armed with this knowledge, we return to the differential equation we were trying to solve earlier, and finally obtain the solution we were expecting. \begin{align} \frac{dy}{dx} &= ky \\ \int k \, dx &= \int\frac{1}{y} \, dy \\ kx + C &= \log(y) \\ y&=e^{kx+C} = e^{kx} \cdot e^{C} = Ae^{kx} \text{ where $A=e^C$} \end{align} Since $a^x$ simply represents $e$ being raised to another base, functions such as $2^x$ are also included in the solution set. In particular, $$ \frac{d}{dx}(a^x) = \frac{d}{dx}(e^{x\log(a)}) = e^{x\log(a)} \cdot \log(a) = a^x \cdot \log(a) \, , $$ and so $k=\log(a)$. Hopefully, this should provide the motivation behind defining the logarithm as $\int_{1}^{x}\frac{1}{t} \, dt$.

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The definite integral $\int_1^x{dt\over t}$ for $x\gt0$ defines a function of some sort. Let's call it $L(x)$, and see what nice properties it enjoys.

Using linearity of integration, we have, for $x,y\gt0$,

$$L(xy)=\int_1^{xy}{dt\over t}=\int_1^x{dt\over t}+\int_x^{xy}{dt\over t}=L(x)+\int_x^{xy}{dt\over t}$$

Now the change of variable $t=xu$ gives us

$$\int_x^{xy}{dt\over t}=\int_1^y{du\over u}=L(y)$$

Thus we have

$$L(xy)=L(x)+L(y)$$

for all $x,y\gt0$, which means $L(x)$ has the defining property of a logarithm. It's not hard to show that $L(x)$ is a strictly increasing continuous function that goes from $0$ at $x=1$ to $\infty$ as $x\to\infty$, hence there is some number, let's call it $e\gt1$, for which $L(e)=\int_1^e{dt\over t}=1$. Because it's "natural" to wonder about the integral of $1/t$, it's natural to call this function the "natural logarithm."

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