A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Restriction for logs
anonymous
 4 years ago
Restriction for logs

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358467325462:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now how do i find the restriction(s) for this ? :S

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1The thing in the bracket may not be less or equal to zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's only supposed to be x>2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont understand why

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1because the x has to satisfy both x>5 and x>2, but since the has to be greater than 2, it can't fall between 5 and 2, because then the equation would be undefined

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1Think about it this way: The two restrictions must be met. Meaning, the number must be greater than 2 AND it must be greater than 5. But, if it is greater than 2, it is greater than 5! So, the greater than 5 is dominated by the greater than 2 restriction. Therefore the only restriction is greater than 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but cant there be more than one restriction ? :S

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1There can. for instance, it it was log(x + 7) + log(x+2), the restrictions would be x + 7 > 0 AND x + 2 > 0 so, it's x< 7 AND x> 2 which reduces to: 2 < x < 7
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.