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anonymous
 3 years ago
Restriction for logs
anonymous
 3 years ago
Restriction for logs

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

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

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

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

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

slaaibak
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0but cant there be more than one restriction ? :S

slaaibak
 3 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
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