Why is $\frac{\log(\frac{1}{x})}{\log x} = -1$ for $x > 1$?

Question

$$\text{For} \ x>1$$ $$\log(1/x) / \log(x) = -1$$ Hi, I noticed this during my studies and I found it kind of weird... I can't find any logarithm rules to see why this happens... I was wondering if maybe I could get some intuition as to why this phenomenon happens.

Answer 1

$\log (\frac 1 x)=\log \, 1- \log \, x=0- \log \, x =- \log \,x$.

Answer 2

We have $$\log\Bigl(\frac{1}{x}\Bigr)=\log(1)-\log(x)=-\log(x)$$ implying your equation.

Answer 3

Note that

$$\frac{1}{x} = x^{-1}$$

Now, what do you know about the logarithm of a power? That is, what can you do with a logarithmic expression of the form $\log(x^a)$ and, hence, $\log(x^{-1})$?

Answer 4

Since $x>1$, we have $\log x \neq 0$. Hence,

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

Answer 5

Try looking at the change of base formula, in general:

$$\frac{\log_c b}{\log_c a} = \frac{\log _c b \cdot \log_a b}{\log_c a\cdot \log _a b} = \frac{\log _c b \cdot \log_a b}{\log_c a^{\log _a b}}=\frac{\log _c b \cdot \log_a b}{\log_c b} = \log_a b$$

Using this formula, we have: $$\frac{\log(\frac{1}{x})}{\log x} = \log_x \bigg(\frac{1}{x}\bigg) = -\log_xx= -1$$