Question

Помогите решить все.

Answer 1

$(frac{1}{49})^{-x}=sqrt{frac{1}{7}}\ (frac{1}{7})^{-2x}=(frac{1}{7})^{frac{1}{2}}\ -2x=frac{1}{2}\ x=-frac{1}{4}$

$4^x+7*2^{x-1}=4,5\ 2^{2x}+7*2^x*2^{-1}=4,5\ t=2^x; t textgreater 0\ t^2+3,5t=4,5\ t^2+3,5t-4,5=0\ D=3,5^2+4*4,5=30,25=5,5^2\ t_1=frac{-3,5+5,5}{2}=1; t_2=frac{-3,5-5,5}{2}=-4,5\ 2^x=1\ x=0$

$frac{1}{3^{5x-3}}leq(frac{1}{3})^{frac{1}{5-3x}}\ (frac{1}{3})^{5x-3}leq(frac{1}{3})^{frac{1}{5-3x}}\ 5x-3leqfrac{1}{5-3x}\ (5x-3)(5-3x)leq1\ 25x-15-15x^2+9x-1leq0\ -15x^2+34x-16leq0\ \ -15x^2+34x-16=0\ D=34^2-4*(-15)*(-16)=196=14^2\ x_1=frac{-34+14}{30}=-frac{2}3; x_2=frac{-34-14}{30}=-1,6\ xin(-infty;-1,6)cup(-frac{2}3;infty)$

$log_2frac{(frac{1}{8})^3*2^{-0,5}}{(frac{1}{4})^2*2^{frac{1}{5}}}=log_2frac{2^{-9}*2^{-0,5}}{2^{-frac{1}{4}}*2^{frac{1}{5}}}=log_2frac{2^{-9,5}}{2^{-0,05}}}=log_22^{-9,55}=-9,55$

$a=log_{frac{1}{5}}frac{7}{5}=log_{frac{1}{5}}7-log_{frac{1}{5}}5^{-1}=log_{frac{1}{5}}7-(-1)=log_{frac{1}{5}}7+1 textgreater 1\ b=(frac{1}{3})^{frac{2}{7}}=frac{1^{frac{2}{7}}}{3^{frac{2}{7}}}=frac{1}{3^{frac{2}{7}}} textless 1\ a textgreater b\ \ \ a=sqrt[4]{10000}=10\ 3^{10} textgreater 500 = textgreater b=log_3500 textless 10\ a textless b \ z$