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This is a somewhat technically specific question about the relationship between $\ln x$ and $e^x$ given one possible definition of $\ln x$.

Suppose that you define $\ln x$ as

$$\ln x = \int_1^x{\frac{dt}{t}}$$

We can use the connection between the integral definition of $\ln x$ and the harmonic series to show that $\ln x$ grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain ($\mathbb{R}$) to its domain ($\mathbb{R^+}$). Let's call that function $\ln^{-1} x$.

Since $\ln x$ grows unboundedly and is zero when $x$ is $1$, we can define $e$ as the unique value such that $\ln e = 1$.

So here's my question: given these starting assumptions, how would you prove that $\ln^{-1} x = e^x$ (or, equivalently, that $\ln x = \log_e x$? I'm having a lot of trouble even seeing how you'd get started proving these facts, since basically every calculus fact I know about $e$ and $\ln x$ presumes this result to be true.

Thanks!

Paramanand Singh
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templatetypedef
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  • Be careful not to use the same variable in both the limits and the integrand, i.e. $\int_1^x \frac{dx}{x}$. You need to use a "dummy variable" in the integrand, e.g. $\int_1^x \frac{dt}{t}$. – Fly by Night Aug 21 '14 at 21:36
  • Because that's how the function is defined...? – Shahar Aug 21 '14 at 21:36
  • @FlybyNight Thanks for fixing that! (Can you tell that it's been a while since I've done any calculus?) – templatetypedef Aug 21 '14 at 21:36
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    @Shahar My confusion is that I can understand that some function has to be in the inverse of $\ln x$, but I don't see why it necessarily has to be $e^x$. That would presume that (1) the inverse of $\ln x$ is some exponential function, and (2) that the base of that exponent is $e$. I'm definitely not doubting that these are true statements, but I don't see how to obtain them. – templatetypedef Aug 21 '14 at 21:37
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    @Shahar That's not very helpful. Most good books on calculus spend several pages connecting the various definitions. – Fly by Night Aug 21 '14 at 21:38
  • @FlybyNight That's why I wrote it as a comment and not an answer. Clearly, I don't quite understand the question. – Shahar Aug 21 '14 at 21:38
  • Can you use $ln$ properties: $ln(x^y) = yln(x)$ or $ln(x/y)=ln(x)-ln(y)$? Not sure if they would help... – Derek Orr Aug 21 '14 at 21:39
  • also, should it not be that $\lim_{x\to 0}\ln x = -\infty$? – Ellya Aug 21 '14 at 21:39
  • @DerekOrr I'd probably need to prove those results first from the definition of $\ln x$. (Sorry if I'm being really conservative in what I'm assuming!) – templatetypedef Aug 21 '14 at 21:40
  • This answer may be of use http://math.stackexchange.com/a/374230/78492 – David Simmons Aug 21 '14 at 21:42
  • Perhaps a good hint would be if two functions are inverses of each other then $f(f^-1(x))=x$. Assume it's $e^x$ and prove the previous equality I mentioned. – Derek Orr Aug 21 '14 at 21:43
  • Sorry I'm bad with syntax ... – Derek Orr Aug 21 '14 at 21:43
  • @Chadman Interesting! That answer works by (at one step) defining $b^x = exp(x \ln b)$. If that's the case, then perhaps there is some element of "it's true because we define it as such." Perhaps it's necessary to rigorously pin down the definition of $e^x$ as well? – templatetypedef Aug 21 '14 at 21:46
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    A similar question has been asked recently. – Lucian Aug 21 '14 at 22:30
  • @FlybyNight That's not actually necessary. The limits are outside the scope of the variable of integration. As such it is formally unambiguous. It is certainly preferable just to avoid confusion, though. – zibadawa timmy Aug 22 '14 at 03:09

3 Answers3

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If $f(x)=\log x$ is defined as a primitive of $\frac{1}{x}$ for which $f(1)=0$, then $f(ab)=f(a)+f(b)$ holds for any couple of positive real numbers. This gives that the inverse function $g(x)$ is a $C^1$ function, satisfies $g(0)=1$ and the functional equation: $$ g(a)g(b)=g(a+b)\tag{1} $$ for any couple of real numbers $a,b$. $(1)$ and differentiability grants that $g'=g$, hence $g(x)=e^x$.

Jack D'Aurizio
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    How did you get from $g' = g$ to $g(x) = e^x$? Clearly this is true, but I'm not sure how you'd prove this without knowing that the derivative of $e^x$ is $e^x$, and (to the best of my knowledge) the proof of that result assumes that $e^x$ is the inverse of $\ln x$. (Or am I mistaken?) – templatetypedef Aug 21 '14 at 21:41
  • $g'=g,g(0)=1$ is a Cauchy problem with a unique solution, and the relation $g(a)g(b)=g(a+b)$ with the conditions $g(0)=1,g\in C^1$ is enough to prove that the exponential function equals its derivative. – Jack D'Aurizio Aug 21 '14 at 21:49
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There have been several questions on definition of $\log x$ and $e^{x}$ and the general power $a^{x}$. Some links are available in various comments on the question. I strongly believe that the real confusion arises because most introductory textbooks on calculus (designed to be consumed by students of 16-17 years of age) don't define any of these functions. They just introduce it with a list of their properties and provide some standard limit formulas. For the benefit of students who are frustrated by such books I have a series of blog posts starting with this one.

The current question defines $\log x$ as an integral and says that $\log e = 1$ and $\log x$ possesses an inverse which is a map from $\mathbb{R}$ to $\mathbb{R}^{+}$. This part is correctly understood by OP. Now the question is how to show that this inverse map is same as $e^{x}, x\in \mathbb{R}$. From the wording of the question it is clear that OP is aware of some definition of $e^{x}$ (although not mentioned in the question) and OP wants to identify it as the inverse of $\log x$. Based on various definitions of $e^{x}$ it is possible to do so (as done in my blog posts). However a nice approach could be to define the expression $e^{x}$ directly as the inverse of $\log x$.

In case one desires to have an independent definition of $e^{x}$ say $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$ then we can show using this definition that $e^{x}$ possesses an inverse and derivative of $y = e^{x}$ is again $y = e^{x}$ so that $dx/dy = 1/y$ and hence $x$ can be defined as the integral $\int (dy/y)$. This way we can identity the inverse of $e^{x}$ with the $\log $ defined as an integral. Same can be done if we define $e^{x}$ as $$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots$$

In any systematic theory of $\log x, e^{x}, a^{x}$ it is sufficient (and perhaps pedagogically better) to define any one of these symbols and develop the rest on the basis of the definition. For example when we start with defining $\log x$ as integral we can define $e^{x}$ as the inverse and $a^{x} = e^{x\log a}$.

Paramanand Singh
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The advantage of defining $\ln x$ as an integral is that it gives a simple way to define $e^x$ in elementary calculus (without resorting to infinite series, taking limits of rational powers of $e$, or other methods).

Once $\ln x$ has been defined in this way, you can establish that $y=\ln x$ is an increasing function whose range is $\mathbb{R}$, so it has an inverse function with domain $\mathbb{R}$ which we can $\mathbf{define}$ to be $y=e^x$.

With this definition, we have the identities

$\ln(e^x)=x$ for all $x$ and $e^{\ln x}=x$ for $x>0$, and $\ln e=1\implies e^{1}=e$.

user84413
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    But then you need to justify the notation $e^x$: you have to prove it's equal to $e$ taken to the power of $x$. – anon Aug 21 '14 at 23:06
  • This is the way many calculus books define $e^x$, so this is what they mean when they take the number e to the power x. This gives probably the simplest way to define what is meant by raising a positive number a to any power x. (Of course, it remains to be shown that this is equivalent to the power series definition of $e^x$, or to the definition that Jack D'Aurizio is using.) – user84413 Aug 21 '14 at 23:11
  • @whacka As the answer indicates, no you don't. It is a definition. Properties of the log give it the expected exponentiation properties, but when introducing log like this those are derived from this definition of $e^x$. That this gives the same definition as other ways of finding $e^x$ is a separate issue to be proven later. – zibadawa timmy Aug 22 '14 at 03:13