ОДЗ : x > 0
1,5\\\\log_{4}^{2}x+\frac{1}{2}log_{4}x-\frac{3}{2}>0\\\\2log_{4}^{2}x+log_{4}x-3>0\\\\log_{4}x=m\\\\2m^{2}+m-3>0\\\\2m^{2}+m-3=0\\\\D=1^{2}-4*2*(-3)=1+24=25=5^{2}\\\\m_{1}=\frac{-1+5}{4}=1\\\\m_{2}=\frac{-1-5}{4}=-\frac{3}{2} \\\\2(m-1)(m+\frac{3}{2})>0\\\\(m-1)(m+1,5)>0" alt="log_{4}^{2}x+log_{4}\sqrt{x}>1,5\\\\log_{4}^{2}x+\frac{1}{2}log_{4}x-\frac{3}{2}>0\\\\2log_{4}^{2}x+log_{4}x-3>0\\\\log_{4}x=m\\\\2m^{2}+m-3>0\\\\2m^{2}+m-3=0\\\\D=1^{2}-4*2*(-3)=1+24=25=5^{2}\\\\m_{1}=\frac{-1+5}{4}=1\\\\m_{2}=\frac{-1-5}{4}=-\frac{3}{2} \\\\2(m-1)(m+\frac{3}{2})>0\\\\(m-1)(m+1,5)>0" align="absmiddle" class="latex-formula">
+ - +
__________₀____________₀__________m
- 1,5 1
\\\\\\\\\\\\\\\\\\\\\ ///////////////////////
1\\\\x>4" alt="1)log_{4} x<-\frac{3}{2}\\\\x<4^{-\frac{3}{2} } \\\\x<(\frac{1}{4})^{\frac{3}{2} }\\\\x<\sqrt{(\frac{1}{4})^{3}}\\\\x<\frac{1}{8}\\\\2)log_{4}x>1\\\\x>4" align="absmiddle" class="latex-formula">
Ответ : x ∈ (0 ; 1/8) ∪ (4 ; + ∞)