When does this logarithmic inequality hold?

Question

Let $a, b$ be integers such that $1<a<b$. When does the following inequality hold? $$\frac{\log b}{\log a}<\log\frac ba$$

Answer 1

Note that

$$\frac {\log b}{\log a}< \log\frac{b}{a}\iff \frac {\log b}{\log a}< \log b - \log a \iff\log b <\log a \log b-\log^2a \\\iff\log^2a-\log a \log\ b+\log b<0$$

and

$$\log a=\frac{\log b\pm\sqrt{\log^2b-4\log b}}{2}$$

therefore

$$\frac{\log b-\sqrt{\log^2b-4\log b}}{2}< \log a<\frac{\log b+\sqrt{\log^2b-4\log b}}{2}$$

with $$\log b\ge 4\implies b\ge e^4$$