Log in and see the ratios!

Let $\log_{2}3 = t$ . Therefore, $\displaystyle \log_{4}3 = \frac{1}{2} .\log_{2}3 = \frac{t}{2}$ . Similarly, $\displaystyle \log_{8}3 = \frac{t}{3}$ .

$\displaystyle \Rightarrow \left( a + t \right) , \left( a+ \frac{t}{2} \right) , \left( a + \frac{t}{3} \right)$ are in geometric progression.

$\displaystyle \Rightarrow \left( a + \frac{t}{2} \right)^2 = \left( a+t \right) \left(a + \frac{t}{3} \right)$

$\displaystyle \therefore a^2 + \frac{t^2}{4} + at = a^2 + \frac{t^2}{3} + \frac{4at}{3} \Rightarrow at - \frac{4at}{3} = \frac{t^2}{3} - \frac{t^2}{4} \Rightarrow - \frac{at}{3} = \frac{t^2}{12}$ .

Since, $t \neq 0$ , $a = - \dfrac{t}{4}$ . Thus the given terms are $\displaystyle \frac{3t}{4} , \frac{t}{4}, \frac{t}{12}$ , which are clearly in a geometric progression with a common ratio of $\boxed{\dfrac{1}{3}}$ .

Assume the common ratio is $q$ , then

$q=\frac{a+\log_{4}{3}}{a+\log_{2}{3}}=\frac{a+\log_{8}{3}}{a+\log_{4}{3}}$ .

Due to the theorem that "if $k=\frac{a}{b}=\frac{c}{d},$ with $b\neq 0,\; d\neq 0,\; b\neq d$ , then $k= \frac{a-c}{b-d}$ " [1] , we have

$q=\frac{(a+\log_{8}{3})-(a+\log_{4}{3})}{(a+\log_{4}{3})-(a+\log_{2}{3})}$

$=\frac{\log_{8}{3}-\log_{4}{3}}{\log_{4}{3}-\log_{2}{3}}$

$=\frac{\log_{2}\sqrt[3]{3}-\log_{2}{\sqrt{3}}}{\log_{2}{\sqrt{3}}-\log_{2}{3}}$

$=\frac{\frac{1}{3}\log_{2}{3}-\frac{1}{2}\log_{2}{3}}{\frac{1}{2}\log_{2}{3}-\log_{2}{3}}$

$=\frac{-\frac{1}{6}\log_{2}{3}}{-\frac{1}{2}\log_{2}{3}}=\frac{1}{3}\approx \boxed{0.333}$ .

[1]. Proof of this theorem:

$\because \frac{a}{b}=\frac{c}{d},$

$\therefore a=bt,\; c=dt,\; q=t$

$\Rightarrow a-c=bt-dt=t(b-d)\Rightarrow t=\frac{a-c}{b-d},$

$\therefore q=\frac{a-c}{b-d}.$