5. Positive real numbers $a, b, c$ form a geometric sequence $(q \neq 1), \log _{a} b, \log _{b} c, \log _{c} a$ form an arithmetic sequence. Then the common difference $d=$
Solution: Let $a=\frac{b}{q}, c=b q, \log _{a} b=x-d, \log _{b} c=x, \log _{c} a=x+d$, then $a^{x-d}=b, b^{x}=c, c^{x+d}=a$. Substituting $a=\frac{b}{q}, c=b q$ yields $\left\{\begin{array}{l}\left(\frac{b}{q}\right)^{x-d}=b \\ b^{x}=b q \\ (b q)^{x+d}=\frac{b}{q}\end{array}\right.$
(2) , Multiplying (1) $\times(3)$ gives $b^{2 x} q^{2 d}=b^{2} q^{-1}$, substituting (2) yields $b^{2} q^{2 d+2}=b^{2} q^{-1}$, thus $q^{2 d+3}=1$, since $q \neq 1$, so $d=-\frac{3}{2}$.