Question

Рівняння - log3(x + 3) - log3(x - 1) = 1 - log3(4 - x)
Повна відповідь

Answer 1

ОДЗ :

$\displaystyle\bf\\\left\{\begin{array}{ccc}x+3 > 0\\x-1 > 0\\4-x > 0\end{array}\right \ \ \ \Rightarrow \ \ \ \left\{\begin{array}{ccc}x > -3\\x > 1\\x < 4\end{array}\right \ \ \ \Rightarrow \ \ \ x\in(1 \ ; \ 4)\\\\\\\log_{3} (x+3)-\log_{3} (x+1)=1-\log_{3} (4-x)\\\\\\\log_{3} (x+3)-\log_{3} (x+1)=\log_{3} 3-\log_{3} (4-x)\\\\\\\log_{3} \frac{x+3}{x+1} =\log_{3} \frac{3}{4-x} \\\\\\\frac{x+3}{x+1} =\frac{3}{4-x} \\\\\\(x+3)\cdot(4-x)=3\cdot(x+1)$

$\displaystyle\bf\\4x-x^{2} +12-3x=3x+3\\\\x-x^{2} +12-3x-3=0\\\\x^{2} +2x-9=0\\\\D=2^{2} -4\cdot(-9)=4+36=40=(2\sqrt{10} )^{2}\\\\\\x_{1} =\frac{-2+2\sqrt{10} }{2} =\sqrt{10} -1\\\\\\x_{2} =\frac{-2-2\sqrt{10} }{2} =-1-\sqrt{10} -ne \ podxodit\\\\\\Otvet \ ; \ \sqrt{10}-1$