It is known that a = ln 3 / 3, B = LN 2 / 2, C = ln 5 / 5 A = (LN2) / 2 = ln (radical 2), B = (Ln3) / 3 = ln (root 3 of degree 3), C = (LN5) / 5 = ln (quintic root 5) Root 5 < root 2 < root 3 So: C

It is known that a = ln 3 / 3, B = LN 2 / 2, C = ln 5 / 5 A = (LN2) / 2 = ln (radical 2), B = (Ln3) / 3 = ln (root 3 of degree 3), C = (LN5) / 5 = ln (quintic root 5) Root 5 < root 2 < root 3 So: C

For example, B-A = (Ln3) / 3 - (LN2) / 2 = (2 * ln3-3 * LN2) / 6 = (Ln3 ^ 2-ln2 ^ 3) / 6 = (ln9-ln8) / 6 = (ln9 / 8) / 6. Because 9 / 8 > 1, B-A > 0. A-c = (LN2) / 2 - (LN5) / 5 = (5 * ln2-2 * LN5) / 10 = (LN2 ^ 5-ln5 ^ 2) / 10 = (ln32-ln25) / 10 = (ln32 / 25) / 10. Because 32 / 25 > 1, a-c > 0. So a > C

Elden Ring Premiere

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