Log2(3x+2)-log2(1-2x)>2
Ответ
1
ОДЗ
![\left \{ {{3x+2>0} \atop {1-2x>0}} \right. <=> \left \{ {{3x>-2} \atop {-2x>-1}} \right. <=>\left \{ {{x>-\frac{2}{3}} \atop {x<\frac{1}{2}}} \right. \\\
(-\frac{2}{3};\frac{1}{2}})](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B3x%2B2%26gt%3B0%7D+%5Catop+%7B1-2x%26gt%3B0%7D%7D+%5Cright.+%26lt%3B%3D%26gt%3B+%5Cleft+%5C%7B+%7B%7B3x%26gt%3B-2%7D+%5Catop+%7B-2x%26gt%3B-1%7D%7D+%5Cright.+%26lt%3B%3D%26gt%3B%5Cleft+%5C%7B+%7B%7Bx%26gt%3B-%5Cfrac%7B2%7D%7B3%7D%7D+%5Catop+%7Bx%26lt%3B%5Cfrac%7B1%7D%7B2%7D%7D%7D+%5Cright.+%5C%5C%5C%0A%28-%5Cfrac%7B2%7D%7B3%7D%3B%5Cfrac%7B1%7D%7B2%7D%7D%29 "\left \{ {{3x+2>0} \atop {1-2x>0}} \right. <=> \left \{ {{3x>-2} \atop {-2x>-1}} \right. <=>\left \{ {{x>-\frac{2}{3}} \atop {x<\frac{1}{2}}} \right. \\\
(-\frac{2}{3};\frac{1}{2}})")
![log_2(3x+2)-log_2(1-2x)>2\\\
log_2\frac{3x+2}{1-2x}>2\\\
\frac{3x+2}{1-2x}>2^2\\\
\frac{3x+2}{1-2x}>4\\\
3x+2<4(1-2x)\\\
3x+2<4-8x\\\
3x+8x<4-2\\\
11x<2\\\
x<\frac{2}{11}](https://tex.z-dn.net/?f=log_2%283x%2B2%29-log_2%281-2x%29%26gt%3B2%5C%5C%5C%0Alog_2%5Cfrac%7B3x%2B2%7D%7B1-2x%7D%26gt%3B2%5C%5C%5C%0A%5Cfrac%7B3x%2B2%7D%7B1-2x%7D%26gt%3B2%5E2%5C%5C%5C%0A%5Cfrac%7B3x%2B2%7D%7B1-2x%7D%26gt%3B4%5C%5C%5C%0A3x%2B2%26lt%3B4%281-2x%29%5C%5C%5C%0A3x%2B2%26lt%3B4-8x%5C%5C%5C%0A3x%2B8x%26lt%3B4-2%5C%5C%5C%0A11x%26lt%3B2%5C%5C%5C%0Ax%26lt%3B%5Cfrac%7B2%7D%7B11%7D "log_2(3x+2)-log_2(1-2x)>2\\\
log_2\frac{3x+2}{1-2x}>2\\\
\frac{3x+2}{1-2x}>2^2\\\
\frac{3x+2}{1-2x}>4\\\
3x+2<4(1-2x)\\\
3x+2<4-8x\\\
3x+8x<4-2\\\
11x<2\\\
x<\frac{2}{11}")
учитывая ОДЗ
учитывая ОДЗ