Why is $\ln\frac{1}{x}=-\ln x$?

Question

I have just discovered this property of the natural logarithm, but why is this statement true? What is the proof that $\ln(1/x)=-\ln(x)$?

No, $x$ has to be positive (assuming complex numbers are not in the game) — imranfat, Mar 05 '18 at 19:24
What is your definition of the logarithm? Is it a real function? complex function? Is it defined as the inverse of the exponential? as the integral of a hyperbola? by a power series? — Xander Henderson, Mar 05 '18 at 19:24
And why can you say that if $e^a= e^b$ then necessarily $a=b$? — Mark Fischler, Mar 05 '18 at 19:26
There are many proofs out there, but as Xander mentions, how do you define the $ln$? Personally, I think that introducing the $ln$ as an integral is a clean approach upon which your question can be very easily answered. — imranfat, Mar 05 '18 at 19:26
Right, because a baby learning to say "Mama" has to be told that some cultures will distinguish cousinhood based an whether the sibling parents are the same gender or not and the baby most specify which culture it is in before we answer the question... $\ln x = k$ means that $e^k = x$ (and it is assumed such a pair of x,k are distinct). So as $e^{-k} = \frac 1{e^k}$ if $x = e^k$ then $e^{-k} = \frac 1x$ so $\ln \frac 1x = -k = -\ln x$. That's it. There's some complication with complex numbers, and there's some complications with what the definition of $e$ is but... let's not worry yet. — fleablood, Mar 05 '18 at 21:30

score 5 · Answer 1 · answered Mar 05 '18 at 19:31

5

Begin with the integral definition of the logarithm function

$$\log(x)\equiv \int_1^x \frac1t\,dt \tag1$$

for $x>0$.

Enforce the substitution $t\mapsto 1/t$ to find

$$\log(x)=-\int_1^{1/x}\frac1t\,dt\tag2$$

Comparing $(1)$ and $(2)$ we find that

$$\log(x)=-\log(1/x)$$

as was to be shown!

answered Mar 05 '18 at 19:31

Mark Viola

179,405

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This is like killing mosquitos with shotguns, don't you think? – cansomeonehelpmeout Mar 05 '18 at 19:33
@cansomeonehelpmeout: why would the definition be a "shotgun"? – Martin Argerami Mar 05 '18 at 19:36
@cansomeonehelpmeout No, I think that it is quite straightforward and elementary. – Mark Viola Mar 05 '18 at 19:37
No, it is the proper way if the $\log(x)$ is defined the way you can see in the first equation. – Botond Mar 05 '18 at 19:37
Why does a minus appear of nowhere when you enforce the substitution t->1/t? And why did you use t instead of x ? – Stallmp Mar 05 '18 at 19:44
Very nice derivation! – user Mar 05 '18 at 19:46
1

When we substitute $t\mapsto 1/t$, we have $dt\mapsto -\frac1{t^2}$. That is the source of the minus sign. And $t$ is a dummy integration variable, while $x$ is an integration limit and fixed. – Mark Viola Mar 05 '18 at 19:51
@gimusi Thank you very much. Much appreciative. – Mark Viola Mar 05 '18 at 19:57

score 3 · Accepted Answer · answered Mar 05 '18 at 19:45

3

By definition

$$y= \ln x \iff e^y=x$$

then

$$y= -\ln x \iff -y= \ln x\iff e^{-y}=x\iff e^y=\frac1x\iff y=\ln \frac1x$$

that is

$$y= -\ln x \iff y=\ln \frac1x$$

therefore

$$-\ln x =\ln \frac1x$$

answered Mar 05 '18 at 19:45

user

154,566

Thanks Mark I really think that all solution given here are of high quality, your way is especially elegant to me! – user Mar 05 '18 at 20:03
You're welcome. My pleasure. – Mark Viola Mar 05 '18 at 20:09

Arthur · Answer 3 · 2018-03-05T19:27:29.247

2

Raise $e$ to the power of $\ln(1/x)$ and to the power of $-\ln x$. See that you get $\frac1x$ either way, and conclude that the numbers must've been equal to begin with (assuming you've already proven that the exponential function is injective).

edited Mar 05 '18 at 19:27

answered Mar 05 '18 at 19:26

Arthur

199,419

Not a proof: $e^a=e^b$ does not automatically imply that $a=b$ – Mark Fischler Mar 05 '18 at 19:27
@MarkFischler I've pointed that out, so I don't see how I've deserved a downvote. – Arthur Mar 05 '18 at 19:30
+1 And actually the existence of an inverse $\ln$ allready indicates that the exponential function is injective. – drhab Mar 05 '18 at 19:35
@drhab We have a square root even though squaring isn't injective, so not really, no. – Arthur Mar 05 '18 at 19:40
A square root is IMHO an inverse of the function $x\mapsto x^2$ with domain $[0,\infty)$ and codomain $[0,\infty)$ (which is injective). – drhab Mar 05 '18 at 19:49
@drhab But there are restrictions on the domain: logarithms are only defined on $(0,\infty)$, so it's even more restricted than the square root. To prove that it really is a full inverse, you can't just say that it is so "because it exists" because that's circular. You can't get away from some kind of proof of the injectivity of $x\mapsto e^x$. Either through calculus and a strictly positive derivative, or injectivity on the rationals plus continuity, or something else. – Arthur Mar 05 '18 at 19:53
But is anything wrong with stating that: $\ln:(0,\infty)\to\mathbb R$ serves as inverse of $\exp:\mathbb R\to(0,\infty)$ hence both functions are injective? – drhab Mar 05 '18 at 19:59
@drhab Not until you've proven that $\exp$ is really injective. Until then that statement is nonsense. – Arthur Mar 05 '18 at 20:09
My answer on the question tells you what I actually had in mind. – drhab Mar 05 '18 at 20:24

cansomeonehelpmeout · Answer 4 · 2018-03-05T19:35:20.037

2

If $x=e^a$ then $a$ is called "$\ln(x)$", the natural log of x. Notice that $$\frac{1}{x}=\frac{1}{e^a}=e^{-a}$$ So $-a=\ln(\frac{1}{x})$ also $-a=-\ln\left (x\right)$. Therefore $\ln\left (\frac{1}{x}\right)=-\ln(x)$.

edited Mar 05 '18 at 19:35

answered Mar 05 '18 at 19:28

cansomeonehelpmeout

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3
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score 0 · Answer 5 · answered Mar 05 '18 at 20:10

The answer to your question depends deeply on your definition of the logarithm function. Here are two possibilities.

Definition: Let $\exp(x) = \mathrm{e}^x$ denote the exponential function. The exponential function is injective (this requires proof), thus it has a well-defined inverse with domain $(0,\infty)$. Define the natural logarithm to be this inverse function. That is, the logarithm is defined by the property that if $a\in\mathbb{R}$ and $b\in (0,\infty)$ and $\exp(a) = b$, then $$ \log(b) := a. $$

Under this definition, \begin{align} \log\left( \frac{1}{x} \right) = y &\iff \mathrm{e}^y = \frac{1}{x} && \text{(defn of $\log$)} \\ &\iff \mathrm{e}^{-y} = x && \text{(properties of $\exp$)} \\ &\iff -y = \log(x) && \text{(defn of $\log$)} \\ &\iff y = -\log(x). && \text{(algebra)} \end{align} We therefore have $$ \log\left( \frac{1}{x} \right) = y = -\log(x), $$ which gives the desired result (via the transitivity of equality).

Definition: For $x > 0$, define the logarithm by $$ \log(x) := \int_{1}^{x} \frac{1}{t}\, \mathrm{d}t. $$

By definition, we have \begin{align} \log\left( \frac{1}{x} \right) &= \int_{1}^{\frac{1}{x}} \frac{1}{t} \,\mathrm{d}t. \end{align} We now want to make the change of variables $u = \frac{1}{t}$. Using the usual computational tricks, we apply the power rule to get $$ \frac{d}u = -\frac{1}{t^2}\,\mathrm{d}t = -u^2\,\mathrm{d}t \implies \mathrm{d}t = -\frac{1}{u^2}\,\mathrm{d}u. $$ Also, when $t = 1$ we have $u=1$, and when $t=\frac{1}{x}$ we have $u = x$ (this deals with the limits of integration). Hence \begin{align} \int_{1}^{\frac{1}{x}} \frac{1}{t} \,\mathrm{d}t &= -\int_{1}^{x} u \cdot \frac{1}{u^2}\,\mathrm{d}u \\ &= -\int_{1}^{x} \frac{1}{u}\,\mathrm{d}u \\ &= -\log(x). \end{align} Putting this all together, we get $$ \log\left( \frac{1}{x} \right) = -\log(x), $$ which is the desired result.

score 0 · Answer 6 · answered Mar 05 '18 at 20:22

It can be shown that function $\exp:\mathbb R\to(0,\infty)$ is bijective or equivalently has an inverse.

In this answer it is preassumed that function $\ln$ is defined as this inverse.

Then for $x>0$ we find:$$\exp(\ln\frac1x)=\frac1x=\frac1{\exp(\ln x)}=\exp(\ln x)^{-1}=\exp(-\ln x)$$ That observation is enough to conclude that: $$\ln\frac1x=-\ln x$$

Why is $\ln\frac{1}{x}=-\ln x$?

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