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Math Help - Finding values of b

  1. #1
    Member mybrohshi5's Avatar
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    Finding values of b

    Not sure if this post should go here so if it would be better else where would a Mod please move this Thanks!

    Here is the question i am working on and it seems simple enough but i just wanted a second opinion

    For what value(s) of b is  b^a > ab for all real numbers a?

    if b is > 2 then  b^a will be > ab

    but

    I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

    Any help on how to prove this problem would be great

    Thank you
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    Not sure if this post should go here so if it would be better else where would a Mod please move this Thanks!

    Here is the question i am working on and it seems simple enough but i just wanted a second opinion

    For what value(s) of b is  b^a > ab for all real numbers a?

    if b is > 2 then  b^a will be > ab

    but

    I also notice that if a = 1 then b can also = 1 so i am not sure what to do with that.

    Any help on how to prove this problem would be great

    Thank you

    It must be b>0 since b<0\Longrightarrow b^a is not defined for many real values of a. If b=0 then the

    inequality is always true insofar we define 0^0=1.

    If a\leq 0 the inequality is trivially true, as the LHS is positive and the RHS is non positive, so we can assume a>0, then:

    b^a\geq ab\Longleftrightarrow a\ln b\geq \ln a+\ln b\Longrightarrow (a-1)\ln b\geq \ln a , and since the ineq. is true for

    a=1 we can assume 0<a\neq 1 , so it must be that

    1) a>1\Longrightarrow \displaystyle{\ln b\geq \frac{\ln a}{a-1}\Longrightarrow b\geq a^\frac{1}{a-1}} .

    But \displaystyle{\lim\limits_{a\to 1^+}a^\frac{1}{a-1}=\lim\limits_{a\to 1^+}e^\frac{\ln a}{a-1}=e} , so it must be b\geq e , and this value already covers the whole

    case a>1 , as \displaystyle{a^\frac{1}{a-1}} has a maximum at a=2

    2) 0<a<1\Longrightarrow we get that \displaystyle{a^\frac{1}{a-1}\xrightarrow [a\to 0]{}\infty} , so in this case b cannot be bounded by some value.

    Tonio
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