Question

Решите пожалуйста 2-ое

Answer 1

$16^{x}-2cdot 12^{x}leq 3^{2x+1}\\16^{x}=(4^2)^{x}=4^{2x}; ; ,; ; 12^{x}=(4cdot 3)^{x}=2^{2x}cdot 3^{x}; ; ,; ; 3^{2x+1}=3^{2x}cdot 3\\4^{2x}-2cdot 4^{x}cdot 3^{x}-3cdot 3^{2x}leq 0; |:3^{2x}\\(frac{4}{3})^{2x}-2cdot (frac{4}{3})^{x}-3leq 0\\t=(frac{4}{3})^{x}>0; ; ,; ; t^2-2t-3leq 0; ; ,; ; t_1=-1; ,; t_2=3; ; (teorema; Vieta)\\(t+1)(t-3)leq 0qquad +++[-1, ]---[, 3, ]+++\\-1leq tleq 3\\-1leq (frac{4}{3})^{x}leq 3$

$0<(frac{4}{3})^{x}leq (frac{4}{3})^{log_{4/3}, 3}; ; ; ; Big [; a^{log_{a}b}=b; ,; ; a>0; ,; ane 1; ,; b>0Big ]\\xleq log_{frac{4}{3}}3\\xleq frac{1}{log_3frac{4}{3}}\\xleq frac{1}{log_34-log_33}\\xleq frac{1}{log_34-1}$