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Math Help - logarithm and domain range finding question

  1. #1
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    logarithm and domain range finding question

    Q: Find the domain of each function.

    g(x)=\log_{5}(\displaystyle{\frac{x+1}{x}})

    now the domain here is x>0 and x<-1

    I want to know how to do calculate that?

    I was thinking / trying to take x to the other side (dividing both sides with x) in this inequality (\displaystyle{\frac{x+1}{x}})>0

    but that removes one x and I can't calculate the domain range as I want it. what I mean is that if I do this (x+1)>0

    then x>-1 which would mean the domain is just greater than -1 which is wrong.

    I hope I explained my question here.

    Thanks in advance.
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  2. #2
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    Re: logarithm and domain range finding question

    \frac{x+1}{x}>0 iff either x + 1 > 0 and x > 0, or x + 1 < 0 and x < 0. In the first case, x + 1 > 0 gives x > -1. However, we need both conditions to be true: x + 1 > 0 and x > 0. The second one is more restrictive (it implies the first one). Therefore, x + 1 > 0 and x > 0 is equivalent to x > 0. Now consider the second case similarly.

    Edit: Dividing both sides of an inequality by x has to be done carefully. If you divide both sides of an equation, you just need to require that x ≠ 0. If you divide both sides of an inequality, you need to consider the cases x > 0 and x < 0 since dividing by a negative number flips the inequality. Thus, one inequality is split into two systems of two inequalities. In this particular problem, it is easier to use the fact that a / b > 0 iff a and b are either both positive or both negative.
    Last edited by emakarov; July 26th 2012 at 10:37 AM.
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  3. #3
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    Re: logarithm and domain range finding question

    Quote Originally Posted by ameerulislam View Post
    Q: Find the domain of each function.

    g(x)=\log_{5}(\displaystyle{\frac{x+1}{x}})

    now the domain here is x>0 and x<-1

    I want to know how to do calculate that?

    I was thinking / trying to take x to the other side (dividing both sides with x) in this inequality (\displaystyle{\frac{x+1}{x}})>0
    You know that {\dfrac{x+1}{x}>0 so the numerator and the denominator must have the same sign.
    That happens if x<-1\text{ or }x>0. Don't over think it.
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  4. #4
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    Re: logarithm and domain range finding question

    Quote Originally Posted by emakarov View Post
    \frac{x+1}{x}>0 iff either x + 1 > 0 and x > 0, or x + 1 < 0 and x < 0. In the first case, x + 1 > 0 gives x > -1. However, we need both conditions to be true: x + 1 > 0 and x > 0. The second one is more restrictive (it implies the first one). Therefore, x + 1 > 0 and x > 0 is equivalent to x > 0. Now consider the second case similarly.

    Edit: Dividing both sides of an inequality by x has to be done carefully. If you divide both sides of an equation, you just need to require that x ≠ 0. If you divide both sides of an inequality, you need to consider the cases x > 0 and x < 0 since dividing by a negative number flips the inequality. Thus, one inequality is split into two systems of two inequalities. In this particular problem, it is easier to use the fact that a / b > 0 iff a and b are either both positive or both negative.
    x>0 is one range (fine with this one) another range is x<-1, considering both numerator and denominator . coz if x is -2 (numerator) than and it divides by -2 (denominator), the result is positive which satisfies logarithms.
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  5. #5
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    Re: logarithm and domain range finding question

    so how to calculate that using considering both conditions.
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  6. #6
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    Re: logarithm and domain range finding question

    Quote Originally Posted by ameerulislam View Post
    so how to calculate that using considering both conditions.
    Quote Originally Posted by emakarov View Post
    Now consider the second case similarly.
    .
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