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Thread: Antilogarithm Inequality

  1. #1

    Antilogarithm Operation on an Inequality

    Here is the inequality

    [TEX]\left( 1 - a \right) < 1[/TEX],

    where [TEX] 0 < a < 1[/TEX]

    Taking the base 2 antilogarithm of both sides gives

    [TEX]2^{\left( 1 - a \right)} < 2[/TEX]

    I am not sure if this is antilogarithm operation is allowed because the quanitity [TEX]\left( 1 - a \right)[/TEX] is less than one.

    P.S. I am not trying to solve this for [TEX]a[/TEX]. I am just using this as an example so I can figure this out.

    Thanks for your help!
    Last edited by onemachine; 02-23-2012 at 02:56 PM.

  2. #2
    Elite Member
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    Quote Originally Posted by onemachine View Post
    Here is the inequality

    [TEX]\left( 1 - a \right) < 1[/TEX],
    where [TEX] 0 < a < 1[/TEX]
    Taking the base 2 antilogarithm of both sides gives
    [TEX]2^{\left( 1 - a \right)} < 2[/TEX]
    I am not sure if this is antilogarithm operation is allowed because the quanitity [TEX]\left( 1 - a \right)[/TEX] is less than one.
    P.S. I am not trying to solve this for [TEX]a[/TEX]. I am just using this as an example so I can figure this out.
    I not sure what antilogarithm means.

    But [TEX]2^x[/TEX] is an everywhere increasing function.

    Therefore for any [TEX]a<b[/TEX] you have [TEX]2^a<2^b[/TEX].
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

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