If $\mathrm{x},\mathrm{y},\mathrm{z}$,are in GP and  ${\mathrm{a}}^{\mathrm{x}}={\mathrm{b}}^{\mathrm{y}}={\mathrm{c}}^{\mathrm{z}}$, then

Given, x, y, z are in GP  and ${\mathrm{a}}^{\mathrm{x}}={\mathrm{b}}^{\mathrm{y}}={\mathrm{c}}^{\mathrm{z}}$ Since,  x, y, z are  in GP.

$\Rightarrow {\mathrm{y}}^2=\mathrm{xz} ......\left(\mathrm{i}\right)$

we have  

${\mathrm{a}}^{\mathrm{x}}={\mathrm{b}}^{\mathrm{y}}={\mathrm{c}}^{\mathrm{z}}=\mathrm{k}\text{ (say); }\mathrm{k}>0$

On taking  log both  sides,  we get

$\begin{array}{r}\mathrm{xlog}\mathrm{a}=\mathrm{ylog}\mathrm{b}=\mathrm{zlog}\mathrm{c}=\log \mathrm{k}\\ \Rightarrow \mathrm{x}=\frac{\log \mathrm{k}}{\log \mathrm{a}},\mathrm{y}=\frac{\log \mathrm{k}}{\log \mathrm{b}},\mathrm{z}=\frac{\log \mathrm{k}}{\log \mathrm{c}}\end{array}$

On substituting  the  values of x,y and  z in Eq.  (i), we  get 

${\left(\frac{\log \mathrm{k}}{\log \mathrm{b}}\right)}^2=\frac{\log \mathrm{k}}{\log \mathrm{a}}\cdot \frac{\log \mathrm{k}}{\log \mathrm{c}}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{(\log \mathrm{k}{)}^2}{(\log \mathrm{b}{)}^2}=\frac{(\log \mathrm{k}{)}^2}{\log \mathrm{a}\cdot \log \mathrm{c}}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\log \mathrm{b}{)}^2=\log \mathrm{a}\cdot \log \mathrm{c}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\log \mathrm{a}}{\log \mathrm{b}}=\frac{\log \mathrm{b}}{\log \mathrm{c}}\end{array}$

$\therefore {\log }_{\mathrm{b}}\mathrm{a}={\log }_{\mathrm{c}}\mathrm{b}\left[\because \frac{{\log }_{\mathrm{e}}\mathrm{x}}{{\log }_{\mathrm{e}}\mathrm{y}}={\log }_{\mathrm{y}}\mathrm{x}\right]$