Log2(x+1)+log2(x+3)=3log3x2-log3x3=-2

Question

Алгебра

Kegrinn

5 год назад

Log2(x+1)+log2(x+3)=3log3x2-log3x3=-2

ОТВЕТЫ

Степан

Nov 20, 2020

$\log_2(x+1)+\log_2(x+3)=3,\quad x \in (-1, + \infty)\\\log_2((x+1)\times(x+3))=3\\\log_2(x^2+3x+x+3)=3\\x^2+3x+x+3=2^3\\x^2+4x+3-8=0\\x^2+5x-x-5=0\\x(x+5)-(x-5) = 0\\(x+5)(x-1)=0\\x+5=0\\x-1=0\\x_1 = -5 \\x_2=1\x \in (-1,+\infty)\\\\\fbox{x=1}$

$\log_3(x^2)-\log_3(x^3)=-2,\quad x\in(0,+\infty)\\2\log_3(x)+3\log_3(x)=-2\\5\log_3(x)=-2\\\log_3(x) = -\dfrac{2}{5}\\x = 3^ {-\dfrac{2}{5}}\\\\ x = \dfrac{\sqrt[5]{3^3}}{3} }=\dfrac{\sqrt[5]{27}}{3} }$

478