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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 $(0,1)$?

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    $\int_1^x\frac{1}{t}dt=-\int_x^1\frac{1}{t}dt$ where $x>0$ – 2017-01-03
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    A Riemann sum approximation for the integral is $\sum_{k=1}^n \frac{1}{t_k}\Delta t$ where $t_k = 1 + \Delta t \frac{k}{n}$ and $\Delta t = \frac{x-1}{n}$. Since $x-1 < 0$ we have $\Delta t < 0$ so loosely speaking we can say that "${\rm d}t < 0$" when $x \in (0,1)$ which explains the sign. – 2017-01-03
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    @Winther, thanks! – 2017-01-03

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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 $(0,1)$.

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    Right, thank you for the comment. However, I am confused because $\frac{1/t}$ is positive on (0,1). Shouldn't it follow then that the area under the curve be positive? – 2017-01-03
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    @JG007 since the integral is from $1$ to $x$ where $x<1$ in this case, the integral does not really just measure the area under the curve. The integral from $x$ to $1$ measures that area. – 2017-01-03
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    @JG007 For instance, $\int_1^2 x dx = \frac{1}{2} 2^2 - \frac{1}{2}1^2 = 3/2$ but $\int_2^1 xdx = \frac{1}{2} 1^2 - \frac{1}{2}2^2 = -3/2.$ – 2017-01-03
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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.