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  3. Mathematics
  4. Statement-1: If log e x > log e y, then 0< y< x. Statement-2: If log a x > log a y, then 0< y< x and 0< a ≠ 1.

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Q. Statement-1: If $\log _e x >\log _e y$, then $0< y< x$.
Statement-2: If $\log _a x >\log _a y$, then $0< y< x$ and $0< a \neq 1$.

Continuity and Differentiability

Solution:

S-1 Clearly, S-1 is true. But S-2 is false.
As, $\log _a x>\log _a y$, then either $0< x< y$ and $0< a< 1$ or $0< y< x$ and $1< a< \infty$.