G1.4 If $\log _{p} x=2, \log _{q} x=3, \log _{r} x=6$ and $\log _{p q r} x=d$, find the value of $d$.
$\begin{array}{l}\frac{\log x}{\log p}=2 ; \frac{\log x}{\log q}=3 ; \frac{\log x}{\log r}=6 \\ \frac{\log p}{\log x}=\frac{1}{2} ; \frac{\log q}{\log x}=\frac{1}{3} ; \frac{\log r}{\log x}=\frac{1}{6} \\ \frac{\log p}{\log x}+\frac{\log q}{\log x}+\frac{\log r}{\log x}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \\ \frac{\log p q r}{\log x}=1 \\ \Rightarrow \frac{\log x}{\log p q r}=1 \\ d=\log _{p q r} x=1\end{array}$