Question

Log2^2(х-1)-log1/2(x-1)>2 помогите решить логарифм пожалуйста

Answer 1

$log_{2} {}^{2} (x - 1) - log_{ \frac{1}{2} }(x - 1) > 2 , \: x > 1\\ log_{2} {}^{2} (x - 1) - log_{ {2}^{ - 1} }(x - 1) - 2 > 0 \\ ( log_{2}(x - 1) ) {}^{2} + log_{2}(x - 1) - 2 > 0 \\ log_{2}(x - 1) = t \\ {t}^{2} + t - 2 > 0 \\ D = {1}^{2} - 4( - 2) = 9 = {3}^{2} \\ t_{1} = \frac{ - 1 + 3}{2} = 1 \\ t _{2} = \frac{ - 1 - 3}{2} = - 2 \\ {t}^{2} + t - 2 = (t - 1)(t + 2) > 0 \\ t \in ( - \infty ; - 2) \cup (1; + \infty ) \\ t < - 2 \\ t > 1$

1)

$log_{2}(x - 1) < - 2 \\ x - 1 < {2}^{ - 2} \\ x < 0.25 + 1 \\ x < 1.25$

2)

$log_{2}(x - 1) > 1 \\ x - 1 > 2 \\ x > 3$

Итого:

$x < 1.25 \\ x > 3 \\ x > 1 \\ x \in (1;1.25) \cup (3; + \infty )$