Since, $a, b$ and $c$ are in $\mathrm{GP}$.
$\therefore b^{2}=a c$
Given equation is
$(\log _{e} \text { a) } x^{2}-\left(2 \log _{e} \text { b) } x+\log _{e} c\right)=0$
Put $x=1,$ we get
$\begin{aligned}
& \log _{e} a-2 \log _{e} b+\log _{e} c=0 \\
\Rightarrow & 2 \log _{e} b=\log _{e} a+\log _{0} c \\
\Rightarrow \quad & \log _{e} b^{2}=\log _{e} a c
\end{aligned}$
$\Rightarrow \quad b^{2}=a c,$ which is true.
Hence, one of the root of given equation is $1 .$
Let another root be $\alpha$.
$\therefore$ Sum of roots, $1+\alpha=\frac{2 \log _{e} b}{\log _{e} a}=\frac{\log _{e} b^{2}}{\log _{e} a}$
$\begin{aligned}
\Rightarrow \quad a &=\frac{\log _{e} a c}{\log _{e} a}-1 \\
&=\frac{\left(\log _{e} a+\log _{e} c\right)}{\log _{e} a}-1 \\
&=\frac{\log _{e} c}{\log _{c} a}=\log _{a} c
\end{aligned}$
Hence, roots are 1 and $\log _{a} c .$