Задача 62849 (x+1) log3(6) +log3(2^x -1/6) <=x-1...
Условие
Решение
[m](x+1)\cdot log_{3}(6) +log_{3}(\frac{2^{x} –1}{6} ≤ (x–1)\cdot log_{3}3[/m]
[m]log_{3}6^{x+1} +log_{3}(\frac{2^{x} –1}{6} ) ≤log_{3}3^{x-1}[/m]
[m]log_{3}6^{x+1}\cdot (\frac{2^{x} –1}{6}) ≤log_{3}3^{x-1}[/m]
[m]6^{x+1}\cdot (\frac{2^{x} –1}{6} ) ≤3^{x-1}[/m]
[m]6^{x}\cdot 6\cdot (\frac{2^{x} –1}{6} ) ≤3^{x-1}[/m]
[m]6^{x}\cdot (2^{x} –1) ≤3^{x-1}[/m]
[m]3^{x}\cdot 2^{x}\cdot (2^{x} –1) ≤3^{x}\cdot \frac{1}{3}[/m]
[m]3\cdot 3^{x}\cdot 2^{x}\cdot (2^{x} –1) ≤3^{x}[/m]
[m]3\cdot 3^{x}\cdot 2^{x}\cdot (2^{x} –1) -3^{x} ≤0[/m]
[m]3^{x}\cdot (3\cdot (2^{x})^2-3 \cdot 2^{x}-1) ≤0[/m]
[m]3^{x}>0[/m] ⇒
[m]3\cdot (2^{x})^2-3 \cdot 2^{x}-1 ≤0[/m]
[m]D=(-3)^2+4\cdot 3\cdot (-1)=9+12=21[/m]
[m]\frac{3-\sqrt{21}}{6} ≤ 2^{x} ≤ \frac{3+\sqrt{21}}{6}[/m]
[m] x ≤ log_{2}\frac{3+\sqrt{21}}{6}[/m]
С учетом ОДЗ получаем о т в е т
[m](1; log_{2}\frac{3+\sqrt{21}}{6}][/m]