Well $\frac{\log{27}}{\log{3}}=3$ and $10$ is clearly much closer to $9$ than it is to $27$, so it is reasonable to assume that we have a number close to $2$. $\frac{22}{8}=2.75$ which is closer to $3$ than it is to $2$. With a bit of thought I could probably find something a little more rigorous, but it didn't seem necessary.Archie, (in effect) you supported that 2 < \(\displaystyle \ \dfrac{log(10)}{log(3)}, \ \ \) and we know that 2 = 16/8 < 22/8.

But I do not see support as to where you show the relative size of \(\displaystyle \ \dfrac{log(10)}{log(3)} \ \) versus 22/8.

E.g: The geometric mean of $9$ and $27$ is $\sqrt{243}$ which is a little larger than $15$. Since $10<15$ we know that $\frac{\log{10}}{\log{3}} < \frac{\log{15}}{\log{3}} < 2.5$

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