Answer :
Following the equation
[tex]V(t) = V_0(1-r)^t[/tex]
We start with an initial price of
[tex]V(0)=V_0[/tex]
and we're looking for a number of years t such that
[tex]V(t)=\dfrac{V_0}{2}[/tex]
If we substitute V(t) with its equation, recalling that
[tex]r = 0.075 \implies 1-r = 0.925[/tex]
we have
[tex]V_0\cdot (0.925)^t=\dfrac{V_0}{2} \iff 0.925^t = \dfrac{1}{2} \iff t = \log_{0.925}\left(\dfrac{1}{2}\right)\approx 8.89[/tex]
So, you have to wait about 9 years.