Log(x) can mean log to the base ten (especially in highschool) or natural log (in college).
So solving for y in log(9/8)=y is just an evaluation of log(9/8).
Either
[tex]y=log_{10}(9/8)=0.05115[/tex] to 5 decimals, or
[tex]y=log_{e}(9/8)=0.11778[/tex] to 5 decimals
depending the context of the question.
