Question

$log_{2}( {x}^{2} - 9x + 8) < 3$
Пожалуйста помогите решить ​

Answer 1

$log_2(x^2-9x+8)<3$

ООФ:

$x^2-9x+8>0\\(x-1)(x-8)>0\\x\in\Big(-\infty;\: 1\Big)\cup\Big(8;\: +\infty\Big)$

$\displaystyle x^2-9x+8<2^3\\x^2-9x<0\\x(x-9)<0\\\\\left \{ {{x<0} \atop {x-9>0}} \right. \\\\\left \{ {{x>0} \atop {x-9<0}} \right. \\\\\left \{ {{x<0} \atop {x>9}} \right. \\\\\left \{ {{x>0} \atop {x<9}} \right. \\\\\\x\in \varnothing \\x\in \Big(0;\:9\Big); x\in\Big(-\infty;\: 1\Big)\cup\Big(8;\: +\infty\Big)$

Ответ: $x\in\Big(0;\: 1\Big)\cup\Big(8;\: 9\Big)$