If you have:
[tex]log_{2}(8)+log_{9}(2)=[/tex]
I would change base into the second log as:
[tex]log_{2}(8)+ \frac{ log_{2}(2)}{ log_{2}(9)}=[/tex]
Solving the logs I get:
[tex]3+ \frac{1}{3} = \frac{10}{3} [/tex]
I used an approx value for the [tex]log_{2}(9)[/tex] using [tex]log_{2}(9)=3[/tex] (it is a little bit more than 3).
