Question

Решите пожалуйста 2-ое

Answer 1

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

$0<(\frac{4}{3})^{x}\leq (\frac{4}{3})^{log_{4/3}\, 3}\; \; \; \; \Big [\; a^{log_{a}b}=b\; ,\; \; a>0\; ,\; a\ne 1\; ,\; b>0\Big ]\\\\x\leq log_{\frac{4}{3}}3\\\\x\leq \frac{1}{log_3\frac{4}{3}}\\\\x\leq \frac{1}{log_34-log_33}\\\\x\leq \frac{1}{log_34-1}$