Equação Logarítmica

por Ludmilla Mayer Sáb 21 Jul 2012, 00:13

log de x na base 2 + log de (2x) na base x = -1

Smile

por danjr5 Sáb 21 Jul 2012, 11:45

Olá Ludmilla,

$log_2 x + log_x 2x = - 1 \\\\ log_2 x + log_x (2.x) = - 1 \\\\ log_2 x + log_x 2 + log_x x = - 1 \\\\ log_2 x + \frac{log_2 2}{log_2 x} + 1 = - 1 \\\\ log_2 x + \frac{1}{log_2 x} + 2 = 0$

Considere

$log_2 x = \lambda$

Seque que

$\lambda + \frac{1}{\lambda} + 2 = 0 \\ \\ \lambda^2 + 2\lambda + 1 = 0 \\ (\lambda + 1)^2 = 0 \\ \\ \lambda + 1 = 0 \\ \lambda = - 1$

Daí,

$log_2 x = \lambda \\\\ log_2 x = - 1 \\\\ 2^{- 1} = x \\\\ {\color{Blue} x = \frac{1}{2}}$