Question

21) log3 x+log3 x-2=0; x+2 x 3 - 4*3 =45; 2x-1 2)(1/3) > = 81; log4(3x+6) < log4 (x+12); 2 3) lg(5lg100) ; -4 -0.25 -0.5 -2 4)0.5 [ (6/5) ] * (0.36) * (0.1);

Answer 1

$log_3^2x+log_3x-2=0, \\ log_3x=t, \\ t^2+t-2=0, \\ t_1=-2, t_2=1, \\ log_3x=-2, \\ x=3^{-2}, \\ x=\frac{1}{9},\\ log_3x=1, \\ x=3.$

$3^{x+2}-4\cdot3^x=45, \\ 3^x(3^2-4)=45, \\ 5\cdot3^x=45, \\ 3^x=9, \\ 3^x=3^2, \\ x=2.$

$(\frac{1}{3})^{2x-1}\geq81, \\ 3^{1-2x}\geq3^4, \\ 1-2x\geq4, \\ -2x\geq3, \\ x\leq-1,5.$

$log_4(3x+6) < log_4(x+12), \\ \begin{cases} 3x+60, \\x+120,\\3x+6-2, \\x2-12,\\x<3; \end{cases} \\ -2<x<3.< var=""$

$lg(5lg100)^2=2lg(5\cdot2lg10)=2lg10=2.$