Question

Решите неравенства:
а). $log ^{2} _{2} x^{2} -15 log_{2}x-4 leq 0$
б). $log ^{2} _{ frac{1}{3} } x^{2} -7 log_{ frac{1}{3} }x+3 leq 0$
в). $log ^{2} _{ 3} x^{2} +13 log_{ 3 }x+3 < 0$
г). $log ^{2} _{ frac{1}{5} } x^{2} -31 log_{ frac{1}{5} }x-8 < 0$

Answer 1

$log ^{2}_{2}x^{2}-15log_{2}x-4 leq 0, \ 2log ^{2}_{2}x-15log_{2}x-4 leq 0,, \ log_{2}x=t, \ 2t^2-15t-4 leq 0, \ 2t^2-15t-4=0, \ D=257>0, \ a=2>0, \ t_1=frac{15-sqrt{257}}{4}, t_2=frac{15+sqrt{257}}{4}, \ frac{15-sqrt{257}}{4} leq t leq frac{15+sqrt{257}}{4}, \ frac{15-sqrt{257}}{4} leq log_{2}x leq frac{15+sqrt{257}}{4}, \ 2^{frac{15-sqrt{257}}{4}} leq x leq 2^{frac{15+sqrt{257}}{4}}.$

$log ^{2}_{frac{1}{3}}x^{2}-7log_{frac{1}{3}}x+3 leq 0, \ 2log ^{2}_{frac{1}{3}}x-7log_{frac{1}{3}}x+3 leq 0, \ log_{frac{1}{3}}x=t, \ 2t^2-7t+3 leq 0, \ 2t^2-7t+3 leq 0, \ D=25=5^2>0, \ a=2>0, \ t_1=frac{1}{2}, t_2=3, \ frac{1}{2} leq t leq 3, \ frac{1}{2} leq log_{frac{1}{3}}x leq 3, \ (frac{1}{3})^{frac{1}{2}} geq x geq (frac{1}{3})^{3}, \ frac{1}{27} leq x leq frac{sqrt{3}}{3}.$

$log ^{2}_{3}x^{2}+13log_{3}x+3 < 0, \ 2log ^{2}_{3}x+13log_{3}x+3 < 0, \ log_{3}x = t, \ 2t^2+13t+3 < 0, \ D=145>0, \ a=2>0, \ t_1=frac{-13-sqrt{145}}{4}, t_2=frac{-13+sqrt{145}}{4}, \ frac{-13-sqrt{145}}{4} < t < frac{-13+sqrt{145}}{4}, \ frac{-13-sqrt{145}}{4} < log_{3}x < frac{-13+sqrt{145}}{4}, \ 3^{frac{-13-sqrt{145}}{4}} < x < 3^{frac{-13+sqrt{145}}{4}}.$


$log^{2}_{frac{1}{5}}x^{2}-31log_{frac{1}{5}}x-8 < 0, \ 2log^{2}_{frac{1}{5}}x-31log_{frac{1}{5}}x-8 < 0, \ log_{frac{1}{5}}x=t, \ 2t^2-31t-8 < 0, \ D=1025=25cdot41>0, \ a=2>0, \ t_1=frac{31-5sqrt{41}}{4}, t_2=frac{31+5sqrt{41}}{4}, \ frac{31-5sqrt{41}}{4} < t < frac{31+5sqrt{41}}{4}, \ frac{31-5sqrt{41}}{4} < log_{frac{1}{5}}x < frac{31+5sqrt{41}}{4}, \ (frac{1}{5})^{frac{31-5sqrt{41}}{4}} > x > (frac{1}{5})^{frac{31+5sqrt{41}}{4}}.$