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Răspuns :

1.

[tex]2log_2^2((2x)^2)-log_2(4\times2x)+\frac{1}{2} =0\\\\4log_2^2(2x)-log_24-log_2(2x)+\frac{1}{2}=0\\\\Notam:log_2(2x)=t\\\\4t^2-2-t+\frac{1}{2}=0\ \ \ |\times2\\\\8t^2-2t-3=0[/tex]

[tex]\Delta=4+96=100\\\\t_1=\frac{2+10}{16} =\frac{3}{4} \\\\t_2=\frac{2-10}{16} =-\frac{1}{2} \\\\\\log_2(2x)=\frac{3}{4}\\\\2x=2^\frac{3}{4}\ \ \ |:2\\\\x=2^-^\frac{1}{4} =\frac{1}{\sqrt[4]{2} }[/tex][tex]log_2(2x)=-\frac{1}{2}\\\\2x=2^-^\frac{1}{2} \ \ \ |:2\\\\x=2^-^\frac{3}{2} \\\\x=\frac{1}{2\sqrt{2} } =\frac{\sqrt{2} }{4}[/tex]

2.

[tex]log_{10}(x^\frac{1}{3} )-3log_{10}^2(x) =0\\\\\frac{1}{3}log_{10}(x)--3log_{10}^2(x) =0\\\\Notam:log_{10}x=t\\\\\frac{1}{3}t-3t^2=0\ \ \ |\times(-3)\\\\9t^2-t=0\\\\t(9t-1)=0\\\\t=0\ si\ t=\frac{1}{9} \\\\log_{10}x=0\\\\x=1\\\\\\log_{10}x=\frac{1}{9}\\\\x=10^\frac{1}{9}[/tex]

Sa mai verifici tu odata calculele. Le-am efectuat direct pe laptop si sa nu fi omis ceva.