Question

sa se demonstreze ca 3^sqrt 5 < 5^sqrt3

Answer 1

3√5<5√3 cum stim ca daca x1<x2 cu x1 si x2 >0 , atuncxi si (x1)² <(x2)², vom ridica la patrta
9*5<25*3
45<75 adevarat

Altfe;
introducem sub radical
 √ (3² *5) <√(5² *3)

√45<√75 adevarat

Answer 2

$\it (3^{\sqrt5})^{\sqrt3} = 3^{\sqrt{15}} \ \textless \ 3^{\sqrt{16}} = 3^4=81 \Longrightarrow (3^{\sqrt5})^{\sqrt3} \ \textless \ 81 \ \ \ (1)$

$\it (5^{\sqrt3})^{\sqrt3} = 5^{\sqrt3 \cdot \sqrt3} = 5^3 = 125 \ \ \ \ (2)$

$\it (1), \ (2) \Longrightarrow (3^{\sqrt5})^{\sqrt3} \ \textless \ 81 \ \textless \ 125 = (5^{\sqrt3})^{\sqrt3} \ \ \ \ (3)$

$\it (3) \Longrightarrow (3^{\sqrt5})^{\sqrt3} \ \textless \ (5^{\sqrt3})^{\sqrt3} \Longrightarrow 3^{\sqrt5} \ \textless \ 5^{\sqrt3} \ \ .$