Interpretation of $\log(x)$ for $x$ on $(0,1)$

Question

If $\log(x) = \int _1 ^x \frac{dt}{t}$ and $\frac{dt}{t}$ is positive on $t\in(0,1)$ but $\log(x)$ is negative on $x \in (0,1)$ then what is the interpretation of $\log(x)$ for $x$ on $(0,1)$?

Answer 1

Same as on $[1,\infty],$ it's just that $\int_1^x\frac{1}{t}dt$ will be a negative number since you are integrating 'backwards' from $1$ down to x when x is in $(0,1)$.

Answer 2

The interpretation of negativeness of $log(x)$ for $x\in(0,1)$ says the point $(x, log x)$ is under the $x$ axis and positiveness of its derivative says that the curve is increasing.