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Math Help - How do you gind domain and range of these logarithmic functions?

  1. #1
    s3a
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    How do you gind domain and range of these logarithmic functions?

    l(x)=log base 5 of (sqrt(x+5))

    AND

    m(x)=log(|x|)

    AND

    n(x)=log((x+1)/(x-1))

    I'd like to find the domain and range WITHOUT needing a graphic calculator. Any help would be appreciated!

    Thanks!
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  2. #2
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    the base has to be positive and different than 1.
    the result has to be positive
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  3. #3
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    Hello s3a
    Quote Originally Posted by s3a View Post
    l(x)=log base 5 of (sqrt(x+5))
    First, note that if \sqrt{x+5} is real x+5 \ge 0.

    So the domain is \{x \in \mathbb{R} : x \ge -5\}

    And since \sqrt{(x+5)} can take any positive value, its log (to any base) can take all values, positive, negative or zero. So the range is the whole of \mathbb{R}.

    Quote Originally Posted by s3a View Post
    m(x)=log(|x|)
    |x| \ge 0 for all x. But \log(|x|) doesn't exist for x = 0, so the domain is \{x \in \mathbb{R} : x \neq 0\}

    The range again is the whole of \mathbb{R}

    Quote Originally Posted by s3a View Post
    n(x)=log((x+1)/(x-1))
    \log\frac{x+1}{x-1} =\log(x+1)- \log(x-1)

    Remember that the input to a log must be >0. Can you see what the domain is?

    And, provided y can take all positive values, \log (y) can take all real values.

    Grandad
    Last edited by Grandad; January 28th 2009 at 10:57 PM. Reason: Submitted in error - corrected version
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  4. #4
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    Quote Originally Posted by s3a View Post
    l(x)=log base 5 of (sqrt(x+5))

    AND

    m(x)=log(|x|)

    AND

    n(x)=log((x+1)/(x-1))

    I'd like to find the domain and range WITHOUT needing a graphic calculator. Any help would be appreciated!

    Thanks!
    If a > 0 and a \neq 1 then the domain of the function f(x)=\log_a(x) is d = \mathbb{R}^+ and the range of the function is r = \mathbb{R}

    to l:

    l(x)=\log_5(\sqrt{x+5})
    Calculating the domain:

    \sqrt{x+5}>0~\implies~x+5>0~\implies~x>-5

    Since \sqrt{x+5} \in (0, +\infty) r = \mathbb{R}

    to m:

    According to my previous considerations you'll get: d = \mathbb{R} \setminus \{0\} and r = \mathbb{R}

    to n:

    n(x)=\log\left(\dfrac{x+1}{x-1}\right)
    Calculating d:

    \dfrac{x+1}{x-1}>0~\implies~\left(x+1>0\wedge x-1>0  \right)~\vee~\left( x+1<0\wedge x-1<0  \right) and therefore:

    d=(-\infty,-1)\cup(1,+\infty)

    If x < -1 then 0 < \dfrac{x+1}{x-1} <1 and therefore n(x) < 0;

    If x > 1 then 1 < \dfrac{x+1}{x-1} and therefore n(x) > 0;

    Since \lim_{|x|\to \infty}\left( \dfrac{x+1}{x-1} \right) = 1 the graph of n approaches the x-axis.

    Therefore r = \mathbb{R}\setminus \{ 0\}
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