Question

Решить логарифмическое неравенство

Answer 1

$4^{log_4^2x}+x^{log_4x}\geq 4\sqrt[4]{4}\; \; ,\; \; ODZ:\; x>0\; ,\; x\ne 1\\\\4^{log_4x\cdot log_4x}+x^{log_4x}\geq 4\sqrt[4]4\\\\(4^{log_4x})^{log_4x}+x^{log_4x}\geq 2^{\frac{5}{2}}\\\\x^{log_4x}+x^{log_4x}\geq 2^{5/2}\; \; ,\; \; \; 2x^{log_4x}\geq 2^{5/2}\; \; ,\; \; \; x^{log_4x}\geq 2^{3/2}\; ,\\\\log_4\Big(x^{log_4x}\Big)\geq lg_42^{3/2}\; \; ,\; \; log_4x\cdot log_4x\geq \frac{3}{2}\cdot \frac{1}{2}\cdot log_22\; \; ,\; \; log^2_4x\geq \frac{3}{4}\; \; ,$

$(log_4x-\frac{\sqrt3}{2})(log_4x+\frac{\sqrt3}{2}) \geq 0\\\\znaki:\; \; +++[-\frac{\sqrt3}{2}\, ]---[\, \frac{\sqrt3}{2}\, ]+++\\\\log_4x\leq -\frac{\sqrt3}{2}\; \; \; ili\; \; \; log_4x\geq \frac{\sqrt3}{2}\\\\log_4x\leq log_4\, 4^{-\sqrt3/2}\; \; \; ili\; \; \; log_4x\geq log_4\, 4^{\sqrt3/2}\\\\0<x\leq \frac{1}{4^{\sqrt3/2}}\; \; \; ili\; \; \; x\geq 4^{\sqrt3/2}\\\\x\in (\, 0\, ;\, \frac{1}{4^{\sqrt3/2}}\; ]\cup [\; 4^{\sqrt3/2}\, ;+\infty )$

Answer 2

Ответ: во вложении Объяснение: