Logarithm Subtraction When I read a textbook of Econometrics, there is derivation about logarithm: $log(x_1) - log(x_0)$ approximately equals to $\frac{(x_1-x_0)} {x_0 }= \frac{\delta x}{x_0}$ . The text noted that you need to use calculus. Could someone can tell how is the result derived? Thank you!
 A: Let's consider the case where positive $\ x\ y\ $ are such that $\ y-x\ $ is relatively small when compared to $\ x,\ $ which means that $\ \frac{y-x}x\ $ is small. Then
$$ \log(y)-\log(x)\ =\ \log(\frac yx)\ =
\ \log\big(1+\frac{y-x}x\big) $$
However, and it's a classic, $\ \log(1+\epsilon)\ $ is approximately equal $\ \epsilon\ $ when $\ \epsilon\ $ is
small. We may consider $\ \frac{y-x}x\ $ to be a small $\ \epsilon.\ $ Then indeed,
$\ \log(y)-\log(x),\ $ i.e.
$\ \log\big(1+\frac{y-x}x\big),\ $ is approximetely
$$ \frac {y-x}x\ =\ \frac{\Delta x}x $$
(Hey, how nice!)

EXTRA:

$$ \forall_{\epsilon>-1}\quad \log(1+\epsilon)\ =
    \ \int_{x=0}^\epsilon\frac {dx}{1+x} $$
Thus, looking at the area under the hyperbole $\ \frac 1{1+\epsilon}\ $ and the respective rectangles, we get
$$ \frac\epsilon{1+\epsilon}\,\ \le\,\ \log(1+\epsilon)
      \,\ \le\,\ \epsilon $$
We get even a sharper inequality by looking at the respective trapezoids:
$$ \frac{2+3\cdot\epsilon}{2+2\cdot\epsilon}\cdot\frac\epsilon{1+\epsilon}
\,\ \le\,\ \log(1+\epsilon)
     \,\ \le\,\ \frac{2+\epsilon}{2+2\cdot\epsilon}\cdot\epsilon $$

REMARK: When $\ -1<\epsilon<0,\ $ you have to concentrate on the above inequalities a bit harder to untangle the negative signs -- the inequalities still hold as written, no big deal. 

