Question

(1/3)^x + 3^x+3 ≤ 12
(34 баллов)

Answer 1

$\displaystyle(\frac{1}{3})^{-x}+3^x+3 \leq 12;\\\frac{1}{3^x}+3^x+3 \leq 12;\ t=3^x;\ t>0;\\\frac{1}{t}+t+3-12\leq 0;$

1+t²-9t ≤ 0;

t²-9t+1=0; решим квадратное уравнение

D=81-4*1=77;

t=(9+√77)/2;

t=(9-√77)/2;

$\displaystyle t=\frac{9\pm\sqrt{77}}{2};\\3^x=\frac{9\pm\sqrt{77}}{2};\\x=log_3\frac{9\pm\sqrt{77}}{2};$

$\displaystyle \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\. \ \ \ \ \ \ \ \ \ log_3\frac{9-\sqrt{77}}{2} \ \ \ \ \ \ \ \ \ \ log_3\frac{9+\sqrt{77}}{2}\\\\x \in [log_3\frac{9-\sqrt{77}}{2};log_3\frac{9+\sqrt{77}}{2}];$