Statement-1: If log e x log e y, then 0

Question

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. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true

Tardigrade Admin · Accepted Answer

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< ∞.

Q. Statement-1: If $lo g_{e} x > lo g_{e} y$ , then $0 < y < x$ .
Statement-2: If $lo g_{a} x > lo g_{a} y$ , then $0 < y < x$ and $0 < a \neq = 1$ .

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is false.

Statement-1 is false, statement-2 is true

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

Q. Statement-1: If loge​x>loge​y, then 0<y<x. Statement-2: If loga​x>loga​y, then 0<y<x and 0<a=1.