I must be missing something. You want to get the logs to the same base. Great, good thinking. Do it
first. But once you have
log5(5), that reduces to 1. Why mess around when that obvious simplification is staring you in the face?
9logx(5)=log5(x)⟹9∗log5(x)log5(5)=log5(x)⟹log5(x)9=log5(x)⟹9={log5(x)}2>0.
Now you could do a u-substitution, but you could directly take the square roots of both sides of the equation. Then what?