a<(b-a)/ln(b/a)<b

$\displaystyle \left(\frac{1}{b}\right)(b-a) < \int_a^b \frac{1}{x} dx < \left(\frac{1}{a}\right)(b-a)$ , for positive $a, b>a$

$\displaystyle\implies \frac{b-a}{b} < \ln(b)-\ln(a) < \frac{b-a}{a}$

$\displaystyle\implies a < \frac{b-a}{\ln(b/a)} < b$

(In fact,

$\displaystyle \int_a^b \frac{1}{x} dx < \left(\frac{1}{b}\right)(b-a)+\left(\frac{1}{2}\right)\left(\frac{1}{a}-\frac{1}{b}\right)(b-a)$

which leads to

$\displaystyle a\left(\frac{2b}{b+a}\right) < \frac{b-a}{\ln(b/a)} < b$ ).