If you have a function in the form of $f(kx)$, the graph is horizontally scaled by a factor of $k$ and
the bigger the magnitude of $k$, the more compressed the graph gets, and the inverse is true.

So by definition, $k$ should be called the horizontal compression factor of the function, meaning if $k = \frac{1}{2}$, the graph is horizontally compressed by a factor of $\frac{1}{2}$, and since stretching is the inverse of compressing you could also say the graph is horizontally stretched by a factor of $2$. and using the same logic, if $k = 2$ then the graph is horizontally compressed by a factor of $2$ or horizontally stretched by a factor of $\frac{1}{2}$

But this is not the case and the accepted practice is to say the graph is compressed by a factor of $k$ if $|k|>1$ and stretched by a factor of $k$ if $0<|k|<1$

This seems extremely unintuitive and to me it doesn't use the word factor correctly. So my question is why do we describe stretching and compressing transformations like this?
['https://math.stackexchange.com/questions/2570174', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/444199/']
The inverse function:

$$ g(y)\ :=\ \frac{\log\_2(y)+1}{\log\_2(y)-1} $$

**REMARK**   Domains:

$$ f : \mathbb R\setminus\{1\}\ \rightarrow\ (0;\infty)\setminus\{2\} $$
and
$$ g : (0;\infty)\setminus{2}\ \rightarrow\ \mathbb R\setminus\{1\} $$

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**Explanation:**

Function $\ f\ $ is a composition of two functions:

$$ f = \psi\circ \phi $$

where $\ \phi(s)\ :=\ \frac{s+1}{s-1}\ $ and $\ \psi(t)\ :=\ 2^t.\ $ The inverse of $\ \phi\ $ is this function itself. The inverse of the exponential function $\ \psi\ $ is $\ \log\_2.\ $ Thus inverse of $\ f\ $ is $\ g\ :=\ \phi\circ\log\_2.$
Answer is :
$$
x = {{\log(y) + \log(2)} \over {\log(y)-\log(2)}}
$$
Demo :   

$$
y = f(x) \\
{{x+1} \over {x-1}} = {{\log y} \over {\log 2}} = P \\
x+1=P(x-1) \\
x+1=Px-P \\
x={{P+1} \over {P-1}} = {{{{\log y} \over {\log 2}}+1} \over {{\log y} \over {\log 2}}-1}
$$