Proving an inequality












1












$begingroup$


Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.



I have no idea how to do this.






inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    I suppose $x ge 1$, and $y$ arbitrarily given.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:28










  • $begingroup$
    @GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
    $endgroup$
    – A. Howells
    Mar 18 '17 at 22:45












  • $begingroup$
    @A.Howells Yes, you're right.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:45
















1












$begingroup$


Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.



I have no idea how to do this.






inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    I suppose $x ge 1$, and $y$ arbitrarily given.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:28










  • $begingroup$
    @GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
    $endgroup$
    – A. Howells
    Mar 18 '17 at 22:45












  • $begingroup$
    @A.Howells Yes, you're right.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:45














1












1








1





$begingroup$


Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.



I have no idea how to do this.






inequality







share|cite|improve this question









$endgroup$




Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.



I have no idea how to do this.






inequality




inequality






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share|cite|improve this question











share|cite|improve this question




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asked Mar 18 '17 at 22:25









pilgrimpilgrim

41828




41828












  • $begingroup$
    I suppose $x ge 1$, and $y$ arbitrarily given.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:28










  • $begingroup$
    @GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
    $endgroup$
    – A. Howells
    Mar 18 '17 at 22:45












  • $begingroup$
    @A.Howells Yes, you're right.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:45


















  • $begingroup$
    I suppose $x ge 1$, and $y$ arbitrarily given.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:28










  • $begingroup$
    @GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
    $endgroup$
    – A. Howells
    Mar 18 '17 at 22:45












  • $begingroup$
    @A.Howells Yes, you're right.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 18 '17 at 22:45
















$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28




$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28












$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45






$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45














$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45




$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45










3 Answers
3






active

oldest

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2












$begingroup$

There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.



This number is the answer.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
    $$
    x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
    $$ similarly,
    $$
    x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
    $$ giving the existence of such an integer $n$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But $x,y$ are positive?! how about $0<x<1$?
      $endgroup$
      – mwomath
      Mar 18 '17 at 23:10










    • $begingroup$
      If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
      $endgroup$
      – Olivier Oloa
      Mar 18 '17 at 23:13










    • $begingroup$
      Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
      $endgroup$
      – mwomath
      Mar 18 '17 at 23:16



















    1












    $begingroup$

    For $x,y>0$ the required inequality is equivalent to write
    $$1le x^{n+frac{1}{2}}yle x.$$
    The right hand side:
    Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
    begin{align}
    x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
    end{align}
    But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
    begin{align}
    x^{n-frac{1}{2}}y le 1
    end{align}
    Taking $ln(cdot)$ for both sides (an increasing function), we get
    begin{align}
    ln x^{n-frac{1}{2}}+ ln y le 0
    end{align}
    i.e.,
    begin{align}
    left{ begin{array}{l}
    nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
    \
    ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
    end{array} right.
    qquad(*)
    end{align}
    So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
    begin{align}
    left{ begin{array}{l}
    nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
    \
    nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
    end{array} right.
    qquad(**)
    end{align}



    Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
    begin{align}
    left{ begin{array}{l}
    left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
    \
    left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
    end{array} right.
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
      $endgroup$
      – Olivier Oloa
      Mar 18 '17 at 23:18












    • $begingroup$
      Yes, I updated my answer. Thank you very much
      $endgroup$
      – mwomath
      Mar 18 '17 at 23:29












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    3 Answers
    3






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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    There is an integer in the interval: $$left[
    log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$    as its diameter is bigger than $1$.



    This number is the answer.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      There is an integer in the interval: $$left[
      log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$      as its diameter is bigger than $1$.



      This number is the answer.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        There is an integer in the interval: $$left[
        log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$        as its diameter is bigger than $1$.



        This number is the answer.






        share|cite|improve this answer











        $endgroup$



        There is an integer in the interval: $$left[
        log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$        as its diameter is bigger than $1$.



        This number is the answer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 6:11


























        community wiki





        2 revs, 2 users 67%
        Hassan
























            1












            $begingroup$

            Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
            $$
            x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
            $$ similarly,
            $$
            x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
            $$ giving the existence of such an integer $n$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But $x,y$ are positive?! how about $0<x<1$?
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:10










            • $begingroup$
              If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:13










            • $begingroup$
              Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:16
















            1












            $begingroup$

            Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
            $$
            x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
            $$ similarly,
            $$
            x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
            $$ giving the existence of such an integer $n$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But $x,y$ are positive?! how about $0<x<1$?
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:10










            • $begingroup$
              If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:13










            • $begingroup$
              Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:16














            1












            1








            1





            $begingroup$

            Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
            $$
            x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
            $$ similarly,
            $$
            x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
            $$ giving the existence of such an integer $n$.






            share|cite|improve this answer









            $endgroup$



            Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
            $$
            x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
            $$ similarly,
            $$
            x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
            $$ giving the existence of such an integer $n$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 18 '17 at 22:37









            Olivier OloaOlivier Oloa

            109k17178294




            109k17178294












            • $begingroup$
              But $x,y$ are positive?! how about $0<x<1$?
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:10










            • $begingroup$
              If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:13










            • $begingroup$
              Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:16


















            • $begingroup$
              But $x,y$ are positive?! how about $0<x<1$?
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:10










            • $begingroup$
              If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:13










            • $begingroup$
              Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:16
















            $begingroup$
            But $x,y$ are positive?! how about $0<x<1$?
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:10




            $begingroup$
            But $x,y$ are positive?! how about $0<x<1$?
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:10












            $begingroup$
            If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
            $endgroup$
            – Olivier Oloa
            Mar 18 '17 at 23:13




            $begingroup$
            If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
            $endgroup$
            – Olivier Oloa
            Mar 18 '17 at 23:13












            $begingroup$
            Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:16




            $begingroup$
            Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:16











            1












            $begingroup$

            For $x,y>0$ the required inequality is equivalent to write
            $$1le x^{n+frac{1}{2}}yle x.$$
            The right hand side:
            Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
            begin{align}
            x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
            end{align}
            But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
            begin{align}
            x^{n-frac{1}{2}}y le 1
            end{align}
            Taking $ln(cdot)$ for both sides (an increasing function), we get
            begin{align}
            ln x^{n-frac{1}{2}}+ ln y le 0
            end{align}
            i.e.,
            begin{align}
            left{ begin{array}{l}
            nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(*)
            end{align}
            So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
            begin{align}
            left{ begin{array}{l}
            nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(**)
            end{align}



            Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
            begin{align}
            left{ begin{array}{l}
            left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
            \
            left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
            end{array} right.
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:18












            • $begingroup$
              Yes, I updated my answer. Thank you very much
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:29
















            1












            $begingroup$

            For $x,y>0$ the required inequality is equivalent to write
            $$1le x^{n+frac{1}{2}}yle x.$$
            The right hand side:
            Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
            begin{align}
            x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
            end{align}
            But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
            begin{align}
            x^{n-frac{1}{2}}y le 1
            end{align}
            Taking $ln(cdot)$ for both sides (an increasing function), we get
            begin{align}
            ln x^{n-frac{1}{2}}+ ln y le 0
            end{align}
            i.e.,
            begin{align}
            left{ begin{array}{l}
            nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(*)
            end{align}
            So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
            begin{align}
            left{ begin{array}{l}
            nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(**)
            end{align}



            Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
            begin{align}
            left{ begin{array}{l}
            left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
            \
            left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
            end{array} right.
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:18












            • $begingroup$
              Yes, I updated my answer. Thank you very much
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:29














            1












            1








            1





            $begingroup$

            For $x,y>0$ the required inequality is equivalent to write
            $$1le x^{n+frac{1}{2}}yle x.$$
            The right hand side:
            Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
            begin{align}
            x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
            end{align}
            But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
            begin{align}
            x^{n-frac{1}{2}}y le 1
            end{align}
            Taking $ln(cdot)$ for both sides (an increasing function), we get
            begin{align}
            ln x^{n-frac{1}{2}}+ ln y le 0
            end{align}
            i.e.,
            begin{align}
            left{ begin{array}{l}
            nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(*)
            end{align}
            So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
            begin{align}
            left{ begin{array}{l}
            nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(**)
            end{align}



            Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
            begin{align}
            left{ begin{array}{l}
            left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
            \
            left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
            end{array} right.
            end{align}






            share|cite|improve this answer











            $endgroup$



            For $x,y>0$ the required inequality is equivalent to write
            $$1le x^{n+frac{1}{2}}yle x.$$
            The right hand side:
            Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
            begin{align}
            x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
            end{align}
            But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
            begin{align}
            x^{n-frac{1}{2}}y le 1
            end{align}
            Taking $ln(cdot)$ for both sides (an increasing function), we get
            begin{align}
            ln x^{n-frac{1}{2}}+ ln y le 0
            end{align}
            i.e.,
            begin{align}
            left{ begin{array}{l}
            nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(*)
            end{align}
            So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
            begin{align}
            left{ begin{array}{l}
            nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
            \
            nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
            end{array} right.
            qquad(**)
            end{align}



            Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
            begin{align}
            left{ begin{array}{l}
            left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
            \
            left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
            end{array} right.
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 18 '17 at 23:28

























            answered Mar 18 '17 at 23:09









            mwomathmwomath

            988614




            988614












            • $begingroup$
              I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:18












            • $begingroup$
              Yes, I updated my answer. Thank you very much
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:29


















            • $begingroup$
              I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
              $endgroup$
              – Olivier Oloa
              Mar 18 '17 at 23:18












            • $begingroup$
              Yes, I updated my answer. Thank you very much
              $endgroup$
              – mwomath
              Mar 18 '17 at 23:29
















            $begingroup$
            I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
            $endgroup$
            – Olivier Oloa
            Mar 18 '17 at 23:18






            $begingroup$
            I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
            $endgroup$
            – Olivier Oloa
            Mar 18 '17 at 23:18














            $begingroup$
            Yes, I updated my answer. Thank you very much
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:29




            $begingroup$
            Yes, I updated my answer. Thank you very much
            $endgroup$
            – mwomath
            Mar 18 '17 at 23:29


















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