4[ Numerical inequalities. Comparisons of numbers.]

What is greater: $\log _{3} 4$ or $\log _{4} 5$ ?
Given the equalities $\log _{3} 4=x$ and $\log _{4} 5=y$, we have: $3^{x}=4,4^{y}=5, x>1, y>1$. Dividing these equalities term by term, we get: $\frac{3^{x}}{4^{y}}=\frac{4}{5}$, which means $\frac{3^{x-1}}{4^{y-1}}=\frac{16}{15}>1$. Therefore, $3^{x-1}>4^{y-1}$. Since $x-1>0$ and $y-$ $1>0$, then $x-1>y-1$, or $x>y$, that is, $\log _{3} 4>\log _{4} 5$.

## Answer

$\log _{3} 4$.