Question

Demonstrati ca numarul a=log in baza9din√3+log in baza 4 din √de ordin 3din 2 este rational

Answer 1

$\it a = log_9\sqrt3+log_4\sqrt[3]2$

Aplicăm formula schimbării de bază :


$\it a = \dfrac{log_3\sqrt3}{log_3 9} +\dfrac{log_2\sqrt[3]2}{log_2 4}= \dfrac{log_3 3^{\frac{1}{2}}}{log_3 3^2} +\dfrac{log_2 2^{\frac{3}{2}}}{log_2 2^2} = \dfrac{\dfrac{1}{2}}{2} + \dfrac{\dfrac{3}{2}}{2} = \dfrac{1}{4} + \dfrac{3}{4} =1$

Deci, a = 1∈ℚ