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 If log3x – log3(x + 1) = 2log33, then x =

Sagot :

kate200468
[tex]1)\ \ \ log_3x-log_3(x + 1) = 2log_33\ \ \ \Rightarrow\ \ \ D:x>0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log_3 \frac{x}{x+1} =log_33^2\ \ \ \Leftrightarrow\ \ \ \ \frac{x}{x+1}=9\ /\cdot(x+1)\\\\x=9(x+1)\\\\x=9x+9\\\\-8x=9\ \ \ \Leftrightarrow\ \ \ x=- \frac{9}{8} \ \notin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\[/tex]

[tex]2)\ \ \ log3x-log3(x + 1) = 2log33\ \ \ \Rightarrow\ \ \ D:x>0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log \frac{3x}{3(x+1)} =log33^2\ \ \ \Leftrightarrow\ \ \ \frac{x}{(x+1)} =1089\ /\cdot(x+1)\\\\x=1089(x+1)\\\\x=1089x+1089\\\\x-1089x=1089\\\\-1088x=1089\ /:(-1088)\\\\x=- \frac{1089}{1088} \ \notin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\Ans.\ the\ equation\ has\ no\ solution.[/tex]
[tex]log3x-log3(x+1)=2log33\ (*)\\\\D:3x > 0\ \wedge\ x+1 > 0\\\\x > 0\ \wedge\ x > -1\\\\D:x\in\mathbb{R^+}\\\\(*)\ log\frac{3x}{3(x+1)}=log33^2\iff \frac{x}{x+1}=1089\\\\1089(x+1)=x\\\\1089x+1089-x=0\\\\1088x=-1089\ \ \ \ /:1088\\\\x=-\frac{1089}{1088}\notin D\\\\Answer:no\ solution;\ x\in\O.[/tex]
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