A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Solve the equations using properties of logs:
a.) 3log[11](x) = 2log[11]8 b.) 2log[5](x) = 3log[5]9
anonymous
 5 years ago
Solve the equations using properties of logs: a.) 3log[11](x) = 2log[11]8 b.) 2log[5](x) = 3log[5]9

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3\log_{11}(x)=2\log_{11}8 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From properties of logs\[3\log_{11}x=\log_{11}x^3\]and\[2\log_{11}8=\log_{11}8^2=\log_{11}64\]so the first equation boils down to\[\log_{11}x^3=\log_{11}64 \rightarrow x^3=64\]so that\[x=4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[B.) 2\log_{5}(x)=3\log_{5}9 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's something similar for the second one too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there were two equations i posted, A and B

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2=9^3=729 \rightarrow x=\pm 27\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I know, but the procedure is the same. The bases are different but it's irrelevant. It's just something to throw you off. As long as the bases are the same in your equation, you can remove the logs in the end using the definition of the log.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the answer for B.) is + or  27?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did the 729 come from?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.