7. Given $\log _{a} x=24, \log _{b} x=40, \log _{\text {abc }} x=$ 
12. Then, $\log _{c} x=$ $\qquad$ .
\begin{array}{l}\text { II.7.60. } \\ \log _{x} c=\log _{x} a b c-\log _{x} a-\log _{x} b \\ =\frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60} .\end{array}