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Is it possible to prove $e^{\ln{x}} = x$ for a student or can you only say that exponentiation is defined to be the inverse of natural logarithm and stop there?

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    $\begingroup$ What are the definitions used? Commonly, the logarithm is defined as the inverse of the exponential function, then it's trivial. It's also trivial if you go the other route and define the logarithm first, and then the exponential function as the inverse of the logarithm. $\endgroup$ – Daniel Fischer Sep 19 '13 at 13:13
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    $\begingroup$ What is your definition of the two functions $\exp$ and $\ln$ ? $\endgroup$ – user37238 Sep 19 '13 at 13:17
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    $\begingroup$ Ah, if the two are independently defined, it is an interesting problem. What were the definitions, and what can one use? One can differentiate, and get $\frac{d}{dx}\left(x\cdot e^{-\ln x}\right) = e^{-\ln x}\cdot\left( 1 - x\cdot \ln' x\right) = 0$, which together with $e^{\ln 1} = 1$ yields the result, if these techniques are available. $\endgroup$ – Daniel Fischer Sep 19 '13 at 13:17
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    $\begingroup$ Maybe one could use properties of logs to show it for rational $x$, and then appeal to continuity, thus avoiding derivatives or other more advanced techniques students may not know. $\endgroup$ – coffeemath Sep 19 '13 at 14:15
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    $\begingroup$ Something to consider is that for negative x, the natural logarithm function isn't defined in the Real number system just as something to be careful. $\endgroup$ – JB King Sep 21 '13 at 3:33
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For a student. Let $x>0$. Then

\begin{align*} e^{ln x} &= x \\ \Leftrightarrow \ln\left(e^{ln x}\right) &= \ln x \\ \Leftrightarrow \ln x \cdot \ln e &= \ln x \\ \Leftrightarrow \ln x \cdot 1 &= \ln x \\ \Leftrightarrow \ln x &= \ln x \\ \Leftrightarrow x &= x. \end{align*}

$ \therefore e^{ln x} = x \, \forall \, x>0.$

Appealing to the fact that $\log a^b=b \cdot \log a$ is perfectly acceptable, as it is a prerequisite for being in the club of people who are allowed to express logarithms.

As my commenters rightfully state, this identity proof is best shown in both directions, with the caveat that $x>0$.

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    $\begingroup$ I like your notation. My question is, is it acceptable to start with what is meant to be proven? Should this proof be upside down instead? $\endgroup$ – jimjim Sep 21 '13 at 3:50
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    $\begingroup$ @user18921 I went with the $\Leftrightarrow$ in lieu of $\Rightarrow$ on your comment. It is actually stronger, and quite true. It is true in either direction, and my implications are not backward in either direction, but this is stronger. $\endgroup$ – J. W. Perry Sep 21 '13 at 4:11
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    $\begingroup$ @Arjang, like J.W. Perry says, either way is good. However, be careful to get the arrows in the correct direction. To prove a statement, we go forwards $(\Rightarrow)$ from the premises and backwards $(\Leftarrow)$ from the conclusion, and we try to meet in the middle. Usually the best way of laying this out is by writing: "Assume [premise 1], [premise 2], etc." Then go forwards from the premises using words like "Thus, therefore" etc. Each time you want to use the conclusion, write "Since we're trying to prove [desired conclusion], thus it suffices to show that [new goal statement]." $\endgroup$ – goblin Sep 21 '13 at 7:53
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    $\begingroup$ But ... How do we know that $\ln e = 1$? Depending on the definitions of $e$ and $\ln$, this is just as non-obvious as the result we're attempting to prove. $\endgroup$ – Blue Sep 21 '13 at 10:40
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    $\begingroup$ @J.W.Perry: If it's as simple as saying that "$\ln$ is the log with base $e$", then the OP's question itself is trivial and your answer unnecessary. (You could simply have cited the log law "$b^{\log_b x} = x$".) Nevertheless, if $e^x$ is defined by, say, the "continuously compounded interest" limit, or by its power series, and if $\ln$ is defined by the integral of the reciprocal function, or by its power series, the connection is far from trivial. $\endgroup$ – Blue Sep 23 '13 at 1:23
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Usually I define the function $\exp(x)$ by being the only continuous function such that $\exp(0)=1$ and $\exp'(x)=\exp(x)$ (it is equal to its derivative).

I define $\log(x)=\int_1^x\frac{dx}{x}$.

Then, I define $f(x)=\exp(\ln(x))$.

By definition, f(1)=1, and $f$ is defined on $(0,\infty)$.

Then I apply the usual definition of derivative.

$f'(x)=\frac{f(x)}{x}$. So $f'(1)=1$.

$f''(x)=\frac{xf'(x)-f(x)}{x^2}=0$.

So $f(1)=f'(1)=1$ and $f''(x)=0$. You deduce that $f(x)=x$.

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  • $\begingroup$ Nice approach! $\;\!$ $\endgroup$ – goblin Sep 21 '13 at 10:04
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There's a variety of good ways to define the exponential function, and infinitely many bad ways. Here's a (necessarily incomplete) list of good ones.

  1. By power series (but you have to prove the series converges). $$\exp : \mathbb{R} \rightarrow \mathbb{R}^+, \;\;\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$

  2. As a solution to an initial value problem (but you have to prove a solution exists). $$\exp : \mathbb{R} \rightarrow \mathbb{R}^+, \;\;\exp(0) = 1,\;\; \exp'(x)=\exp(x)$$

  3. Via a well-known limit (but you have to prove convergence). $$\exp : \mathbb{R} \rightarrow \mathbb{R}^+, \;\; \exp(x) = \mathrm{lim}_{n \rightarrow \infty} \left(1+\frac{x}{n}\right)^n$$

  4. As a function that is undone by the logarithm (but you have to prove that there exists a unique function with this property, or in other words that the logarithm is invertible). $$\exp : \mathbb{R} \rightarrow \mathbb{R}^+,\;\;\log(\exp(x)) = x$$

Note that in each case, we're implicitly asserting that the equation of interest holds for all $x$ in the domain of $\exp$, namely $\mathbb{R}.$

Personally, I like definition 4 best, because the others are "too clever." Math is best when there's no "magic," and the reader ends up feeling like they could have worked it all out themselves, had they just the time and the motivation. So, we must define the (natural) logarithm first. As with the exponential function, there's a variety of good ways of doing this, and infinitely many bad ways. Here's the only good way that I know of.

$$\log : \mathbb{R}^+ \rightarrow \mathbb{R},\;\;\log(x) = \int_1^x\frac{dy}{y}$$

Now, it is necessary to prove that this integral exists. However, if $1/y$ is known to be continuous on $\mathbb{R}^+,$ then existence is immediate.

Furthermore, it is necessary to prove invertibility. However, since $1/y$ is strictly positive on $\mathbb{R}^+$, thus $\log$ is a strictly increasing function. Therefore, $\log$ is injective (or 'one-to-one'). So all that remains to show is that $\log$ is surjective (or 'onto'). Once we have surjectivity, it is immediate that $\log$ has an inverse, which we call $\exp$.

So, if we do things this way, it is not necessary to prove that $\exp(\log x) = x$, because it is our very definition.

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    $\begingroup$ What about the approach of defining $b^x = \sup \{ b^q \mid q \in \mathbb Q, q \leq x \}$ (for $b > 1$), proving that $\frac{d}{dx} b^x = c b^x$ for some constant $c$ that depends on $b$, and then defining $e$ to be the particular choice of $b$ such that this constant is equal to $1$. Is there a difficulty with that approach? I would guess that this line of reasoning is what led to the discovery of $e$ in the first place. And this approach seems less clever than definition 4. $\endgroup$ – littleO Sep 21 '13 at 7:58
  • $\begingroup$ @littleO, I love it! You should write a short paper on it. $\endgroup$ – goblin Sep 21 '13 at 8:02
  • $\begingroup$ One big reason to prefer the power series definition is that you get the definitions for $\sin$ and $\cos$ as well as Euler's formula more or less for free. Of course this doesn't work in a secondary education environment because proving results about power series (such as "term-by-term differentiation and integration works") requires a fairly complete theory of uniform convergence. $\endgroup$ – kahen Sep 21 '13 at 8:54
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For $b > 1$ and $x>0$, $\log_b(x)$ is defined to be the number $z$ such that $b^z = x$.

$\log_b(x)$ could be called "the exponent from $b$ to $x$".

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  • $\begingroup$ This is the same as defining $\log_b(y)$ to be the inverse of $b^x$. $\endgroup$ – Josephine Moeller Sep 21 '13 at 5:03
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Another way to show (but probably not rigorously prove) $e^{\ln{x}}=x$ is to use a linear differential equation of the first order: $\frac{dy}{dx}+y \ P(x) \ = \ Q(x) \quad \Rightarrow \quad dy \ + y \ P(x) \ dx \ = Q(x) \ dx$.

First, let $\mu(x)=e^{\ln{x}}=e^{\int{P(x)dx}} \enspace \Rightarrow \enspace \int{P(x) \ dx}=\ln{x} \enspace \Rightarrow \enspace \frac{d}{dx}\int{P(x) \ dx}=\frac{d}{dx}\ln{x}$ $$\Rightarrow \enspace P(x) \ dx=\frac{1}{x} \ dx \enspace \Rightarrow \enspace P(x) =\frac{1}{x}$$

Now, multiply both sides of the differential equation by the integrating factor $\mu(x)$: $\mu(x)[dy \ + y \ P(x) \ dx] \ = \mu(x) \ Q(x) \ dx$.
If we let $Q(x)=0$, the equation becomes homogeneous: $\mu(x)[dy \ + y \ P(x) \ dx] \ = 0$.

Remembering that $P(x)=\frac{1}{x}$, this differential equation can be solved two ways, first using $\mu(x)=e^{\ln{x}}$: $$e^{\ln{x}}[dy \ + y \ P(x) \ dx] \ = 0$$ $$e^{\ln{x}}[dy \ + y \ \frac{1}{x} \ dx] \ = 0 \quad (A)$$

Alternatively, we could use $\mu(x)=x$ (if $e^{\ln{x}}=x$): $$x[dy \ + y \ P(x) \ dx] \ = 0$$ $$x[dy \ + y \ \frac{1}{x} \ dx] \ = 0$$ $$x \ dy + y \ dx \ = 0 \quad (B)$$

Set $(A) = (B)$: $$e^{\ln{x}}[dy \ + y \ \frac{1}{x} \ dx] \ = x \ dy + y \ dx$$

Integrate each side: $$ye^{\ln{x}} + c = yx + c$$ If we can assume the $c$'s are the same, $$ye^{\ln{x}}=yx$$ $$e^{\ln{x}}=x$$

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Well, it is a property that follows from the logarithm's definition.

If $log_ab =c$ with the condition $a \in \mathbb{R^+} - \{0\}$

You know this logarithm can be writte as $a^c =b$ and is obviously that $log_ab = log_ab$, i.e is equal to itself. Using the property $a^c=b$ you can see that $a^{log_ab} = b$, then replace $a$ by euler number $e$,to get:

$e^{log_eb}=b \to e^{lnb}=b$

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$$e^{\ln x}=x$$ $$\ln (e^{\ln x})=\ln x$$ $$(\ln e)\cdot(\ln x)=\ln x$$ So since $\ln e=1,$ $$\ln x=\ln x$$ then $x=x.$

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    $\begingroup$ This isn't really a proof. It assumes that $e^{\ln x}=x$ in the outset, which is what is to be proved. $\endgroup$ – Cameron Buie Sep 21 '13 at 4:34
  • $\begingroup$ To make this a legitimate proof, just show that each statement is true if and only if the previous one is true. Then since you reach a trivial conclusion, you can just go backwards and argue that your last equality implies the first. $\endgroup$ – YiFan Jul 30 '18 at 5:47

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