Get expert insights and reliable answers to your questions on IDNLearn.com. Find the solutions you need quickly and accurately with help from our knowledgeable community.
Sagot :
[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]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]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.
