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 $\ln(1/x)=-\ln(x)$?

Answer 1

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

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

Answer 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).

Answer 3

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 $\ln\left (\frac{1}{x}\right)=-\ln(x)$.

Answer 4

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!

Answer 5

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).

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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.

Answer 6

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