I haven't seen this typesetting before, but I'm assuming we want
$\displaystyle \left(2^{\log_26} \right)\left(3^{\log_95} \right)\left(5^{\log_a2} \right)$
(Edited)
The middle factor can be evaluated as follows.
$\displaystyle 3^{\log_95}=3^{\frac{\log_35}{\log_39}}=3^{\frac{1 }{2}\cdot\log_35}=\left( 3^{\log_35}\right)^{\left(\frac{1}{2}\right)}=\sqr t{5}$