Question

Решить логарифмическое неравенство

Answer 1

$4^{log_4^2x}+x^{log_4x}geq 4sqrt[4]{4}; ; ,; ; ODZ:; x>0; ,; xne 1\\4^{log_4xcdot log_4x}+x^{log_4x}geq 4sqrt[4]4\\(4^{log_4x})^{log_4x}+x^{log_4x}geq 2^{frac{5}{2}}\\x^{log_4x}+x^{log_4x}geq 2^{5/2}; ; ,; ; ; 2x^{log_4x}geq 2^{5/2}; ; ,; ; ; x^{log_4x}geq 2^{3/2}; ,\\log_4Big(x^{log_4x}Big)geq lg_42^{3/2}; ; ,; ; log_4xcdot log_4xgeq frac{3}{2}cdot frac{1}{2}cdot log_22; ; ,; ; log^2_4xgeq frac{3}{4}; ; ,$

$(log_4x-frac{sqrt3}{2})(log_4x+frac{sqrt3}{2}) geq 0\\znaki:; ; +++[-frac{sqrt3}{2}, ]---[, frac{sqrt3}{2}, ]+++\\log_4xleq -frac{sqrt3}{2}; ; ; ili; ; ; log_4xgeq frac{sqrt3}{2}\\log_4xleq log_4, 4^{-sqrt3/2}; ; ; ili; ; ; log_4xgeq log_4, 4^{sqrt3/2}\\0<xleq frac{1}{4^{sqrt3/2}}; ; ; ili; ; ; xgeq 4^{sqrt3/2}\\xin (, 0, ;, frac{1}{4^{sqrt3/2}}; ]cup [; 4^{sqrt3/2}, ;+infty )$

Answer 2

Ответ: во вложении Объяснение: