Logeometric progression

I came at it from a slightly different angle.

Since the three terms are in G.P. we can write

$r(a+\log_2 3)=a+\log_4 3$

$r(a+\log_4 3)=a+\log_8 3$

Subtracting the equations eliminates a to give

$r(\log_2 3-\log_4 3)=\log_4 3-\log_8 3$

Now change all the logarithms to base 2 to get

$r\left(\log_2 3-\dfrac{\log_2 3}{2}\right)=\dfrac{\log_2 3}{2}-\dfrac{\log_2 3}{3}$

Now the logarithm can be cancelled as a factor leaving

$\dfrac{r}{2}=\dfrac{1}{6}$

and so

$\boxed{r=\dfrac{1}3{}}$

Let $x = \log_{2}(3).$ Then by the change of base rule the progression becomes

$a + x, a + \dfrac{x}{2}, a + \dfrac{x}{3}.$

As the progression is geometric, we must have that

$\dfrac{a + \dfrac{x}{2}}{a + x} = \dfrac{a + \dfrac{x}{3}}{a + \dfrac{x}{2}} \Longrightarrow a^{2} + ax + \dfrac{x^{2}}{4} = a^{2} + \dfrac{4ax}{3} + \dfrac{x^{2}}{3}$

$\Longrightarrow -\dfrac{ax}{3} = \dfrac{x^{2}}{12} \Longrightarrow -4ax = x^{2}.$

Now clearly $x \ne 0,$ so $a = -\dfrac{x}{4}.$ The desired ratio is then

$\dfrac{-\dfrac{x}{4} + \dfrac{x}{2}}{-\dfrac{x}{4} + x} = \dfrac{\dfrac{x}{4}}{\dfrac{3x}{4}} = \dfrac{1}{3} = \boxed{0.333}$ to 3 decimal places.