27.24. Compare the numbers $\log _{3} 7$ and $\log _{7} 27$.

## 27.5. Irrationality of logarithms
27.24. It is clear that $\log _{7} 27=3 \log _{7} 3=\frac{3}{\log _{3} 7}$. Let's prove the inequality $\left(\log _{3} 7\right)^{2}>3$, i.e., $\log _{3} 7>\sqrt{3}$. It is easy to verify that $\sqrt{3}3^{\sqrt{3}}$ follows from the inequality $7^{4}>3^{7}$, i.e., $2401>$ $>2187$. In the end, we get that $\log _{3} 7>\log _{7} 27$.