ĐK: $0<x\ne 1; a>0; b>0$
$\log_a x=7$
$\log_b x=3$
$\log_{ab}x=\dfrac{\log_a x}{\log_a (ab)}$
$=\dfrac{\log_a x}{1+\log_a b}$
Mà $\log_b x=3\to \log_x b=\dfrac{1}{3}$
$\to \log_a b=\log_a x.log_x b=\dfrac{7}{3}$
Vậy:
$\log_{ab}x=\dfrac{7}{1+\dfrac{7}{3}}=\dfrac{21}{10}$