Question

If log3x – log3(x + 1) = 2log3 3, then x =

Answer 1

`log3x-log3(x+1)=2log33\ (*)\\\\D:3x > 0\ \wedge\ x+1 > 0\\\\x > 0\ \wedge\ x > -1\\\\D:x\in\mathbb{R^+}\\\\(*)\ log(3x)/(3(x+1))=log33^2\iff (x)/(x+1)=1089\\\\1089(x+1)=x\\\\1089x+1089-x=0\\\\1088x=-1089\ \ \ \ /:1088\\\\x=-(1089)/(1088)\\otin D\\\\Answer:no\ solution;\ x\in\O.`

Answer 2

`1)\ \ \ log_3x-log_3(x + 1) = 2log_33\ \ \ \Rightarrow\ \ \ D:x>0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log_3 (x)/(x+1) =log_33^2\ \ \ \Leftrightarrow\ \ \ \ (x)/(x+1)=9\ /\cdot(x+1)\\\\x=9(x+1)\\\\x=9x+9\\\\-8x=9\ \ \ \Leftrightarrow\ \ \ x=- (9)/(8) \ \\otin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\`


`2)\ \ \ log3x-log3(x + 1) = 2log33\ \ \ \Rightarrow\ \ \ D:x>0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log (3x)/(3(x+1)) =log33^2\ \ \ \Leftrightarrow\ \ \ (x)/((x+1)) =1089\ /\cdot(x+1)\\\\x=1089(x+1)\\\\x=1089x+1089\\\\x-1089x=1089\\\\-1088x=1089\ /:(-1088)\\\\x=- (1089)/(1088) \ \\otin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\Ans.\ the\ equation\ has\ no\ solution.`