How can I compare logarithm and the number? [duplicate]












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This question already has an answer here:




  • Compare $log_34$ and $2^frac 1 4$

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There are two numbers $log_3 4$ and $sqrt[4]{2}$. How they can be compared without calculator?






logarithms number-comparison







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marked as duplicate by José Carlos Santos, Thomas Andrews, David K, max_zorn, Jyrki Lahtonen Jan 31 at 21:23


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 6




    $begingroup$
    Possible duplicate of Compare $log_34$ and $2^frac 1 4$
    $endgroup$
    – Flame Trap
    Jan 31 at 19:08








  • 3




    $begingroup$
    Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
    $endgroup$
    – Thomas Andrews
    Jan 31 at 19:09
















0












$begingroup$



This question already has an answer here:




  • Compare $log_34$ and $2^frac 1 4$

    1 answer




There are two numbers $log_3 4$ and $sqrt[4]{2}$. How they can be compared without calculator?






logarithms number-comparison







share|cite|improve this question









$endgroup$



marked as duplicate by José Carlos Santos, Thomas Andrews, David K, max_zorn, Jyrki Lahtonen Jan 31 at 21:23


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 6




    $begingroup$
    Possible duplicate of Compare $log_34$ and $2^frac 1 4$
    $endgroup$
    – Flame Trap
    Jan 31 at 19:08








  • 3




    $begingroup$
    Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
    $endgroup$
    – Thomas Andrews
    Jan 31 at 19:09














0












0








0


1



$begingroup$



This question already has an answer here:




  • Compare $log_34$ and $2^frac 1 4$

    1 answer




There are two numbers $log_3 4$ and $sqrt[4]{2}$. How they can be compared without calculator?






logarithms number-comparison







share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Compare $log_34$ and $2^frac 1 4$

    1 answer




There are two numbers $log_3 4$ and $sqrt[4]{2}$. How they can be compared without calculator?





This question already has an answer here:




  • Compare $log_34$ and $2^frac 1 4$

    1 answer







logarithms number-comparison




logarithms number-comparison






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 31 at 19:02









Totum_RevolutumTotum_Revolutum

31




31




marked as duplicate by José Carlos Santos, Thomas Andrews, David K, max_zorn, Jyrki Lahtonen Jan 31 at 21:23


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by José Carlos Santos, Thomas Andrews, David K, max_zorn, Jyrki Lahtonen Jan 31 at 21:23


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 6




    $begingroup$
    Possible duplicate of Compare $log_34$ and $2^frac 1 4$
    $endgroup$
    – Flame Trap
    Jan 31 at 19:08








  • 3




    $begingroup$
    Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
    $endgroup$
    – Thomas Andrews
    Jan 31 at 19:09














  • 6




    $begingroup$
    Possible duplicate of Compare $log_34$ and $2^frac 1 4$
    $endgroup$
    – Flame Trap
    Jan 31 at 19:08








  • 3




    $begingroup$
    Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
    $endgroup$
    – Thomas Andrews
    Jan 31 at 19:09








6




6




$begingroup$
Possible duplicate of Compare $log_34$ and $2^frac 1 4$
$endgroup$
– Flame Trap
Jan 31 at 19:08






$begingroup$
Possible duplicate of Compare $log_34$ and $2^frac 1 4$
$endgroup$
– Flame Trap
Jan 31 at 19:08






3




3




$begingroup$
Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:09




$begingroup$
Find a rational number between them. In this case you can use $frac{5}4.$ Since $3^5=243<256=4^4$ you have $3^{5/4}<4$ and thus $log_3 4 > frac{5}{4}.$ Do the same to show that $sqrt[4]{2}<frac{5}{4}.$
$endgroup$
– Thomas Andrews
Jan 31 at 19:09










1 Answer
1






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oldest

votes


















0












$begingroup$

It's not trivial in general case. But for your expression:
$$
I=frac{sqrt[4]{2}}{log_3 4}=frac{a}{log_3 a^{8}}=frac18frac{a}{log_3 a}=f(a).
$$



where $a=2^{1/4}$.
Function $f(x)=x/(8log_3x)=frac{ln 3}8 x/ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/ln x$).



Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable):
$$ 3^{2/13} quad ? quad 2^{1/4}\
3^8quad ? quad2^{13}\
6561 quad < quad 8192$$
then we calculate $f(b)$:
$$
f(b) = frac{3^{2/13}}{8log_33^{2/13}}= frac{13}{16} 3^{2/13}.
$$
Then we show that $f(b) < 1$ (still doable):
$$
13cdot3^{2/13} quad ?< quad 16\
13^{13}3^2 quad ?< quad 16^{13}\
13^{12}3^2 quad ?< quad 16^{12}\
13^{6}3 quad ?< quad 16^{6}=2^{24}\
text{since} 13^23=507<512=2^9\
13^{4} quad ?< quad 2^{15}qquad \28561 quad < quad 32678
$$



Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    It's not trivial in general case. But for your expression:
    $$
    I=frac{sqrt[4]{2}}{log_3 4}=frac{a}{log_3 a^{8}}=frac18frac{a}{log_3 a}=f(a).
    $$



    where $a=2^{1/4}$.
    Function $f(x)=x/(8log_3x)=frac{ln 3}8 x/ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/ln x$).



    Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable):
    $$ 3^{2/13} quad ? quad 2^{1/4}\
    3^8quad ? quad2^{13}\
    6561 quad < quad 8192$$
    then we calculate $f(b)$:
    $$
    f(b) = frac{3^{2/13}}{8log_33^{2/13}}= frac{13}{16} 3^{2/13}.
    $$
    Then we show that $f(b) < 1$ (still doable):
    $$
    13cdot3^{2/13} quad ?< quad 16\
    13^{13}3^2 quad ?< quad 16^{13}\
    13^{12}3^2 quad ?< quad 16^{12}\
    13^{6}3 quad ?< quad 16^{6}=2^{24}\
    text{since} 13^23=507<512=2^9\
    13^{4} quad ?< quad 2^{15}qquad \28561 quad < quad 32678
    $$



    Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It's not trivial in general case. But for your expression:
      $$
      I=frac{sqrt[4]{2}}{log_3 4}=frac{a}{log_3 a^{8}}=frac18frac{a}{log_3 a}=f(a).
      $$



      where $a=2^{1/4}$.
      Function $f(x)=x/(8log_3x)=frac{ln 3}8 x/ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/ln x$).



      Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable):
      $$ 3^{2/13} quad ? quad 2^{1/4}\
      3^8quad ? quad2^{13}\
      6561 quad < quad 8192$$
      then we calculate $f(b)$:
      $$
      f(b) = frac{3^{2/13}}{8log_33^{2/13}}= frac{13}{16} 3^{2/13}.
      $$
      Then we show that $f(b) < 1$ (still doable):
      $$
      13cdot3^{2/13} quad ?< quad 16\
      13^{13}3^2 quad ?< quad 16^{13}\
      13^{12}3^2 quad ?< quad 16^{12}\
      13^{6}3 quad ?< quad 16^{6}=2^{24}\
      text{since} 13^23=507<512=2^9\
      13^{4} quad ?< quad 2^{15}qquad \28561 quad < quad 32678
      $$



      Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It's not trivial in general case. But for your expression:
        $$
        I=frac{sqrt[4]{2}}{log_3 4}=frac{a}{log_3 a^{8}}=frac18frac{a}{log_3 a}=f(a).
        $$



        where $a=2^{1/4}$.
        Function $f(x)=x/(8log_3x)=frac{ln 3}8 x/ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/ln x$).



        Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable):
        $$ 3^{2/13} quad ? quad 2^{1/4}\
        3^8quad ? quad2^{13}\
        6561 quad < quad 8192$$
        then we calculate $f(b)$:
        $$
        f(b) = frac{3^{2/13}}{8log_33^{2/13}}= frac{13}{16} 3^{2/13}.
        $$
        Then we show that $f(b) < 1$ (still doable):
        $$
        13cdot3^{2/13} quad ?< quad 16\
        13^{13}3^2 quad ?< quad 16^{13}\
        13^{12}3^2 quad ?< quad 16^{12}\
        13^{6}3 quad ?< quad 16^{6}=2^{24}\
        text{since} 13^23=507<512=2^9\
        13^{4} quad ?< quad 2^{15}qquad \28561 quad < quad 32678
        $$



        Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.






        share|cite|improve this answer









        $endgroup$



        It's not trivial in general case. But for your expression:
        $$
        I=frac{sqrt[4]{2}}{log_3 4}=frac{a}{log_3 a^{8}}=frac18frac{a}{log_3 a}=f(a).
        $$



        where $a=2^{1/4}$.
        Function $f(x)=x/(8log_3x)=frac{ln 3}8 x/ln x$ is decreasing from $1$ to $e$ (one can show it by taking derivative of $x/ln x$).



        Let's take $b=3^{2/13}$. First, we show that $b<a$ (it's not easy without calculator, but doable):
        $$ 3^{2/13} quad ? quad 2^{1/4}\
        3^8quad ? quad2^{13}\
        6561 quad < quad 8192$$
        then we calculate $f(b)$:
        $$
        f(b) = frac{3^{2/13}}{8log_33^{2/13}}= frac{13}{16} 3^{2/13}.
        $$
        Then we show that $f(b) < 1$ (still doable):
        $$
        13cdot3^{2/13} quad ?< quad 16\
        13^{13}3^2 quad ?< quad 16^{13}\
        13^{12}3^2 quad ?< quad 16^{12}\
        13^{6}3 quad ?< quad 16^{6}=2^{24}\
        text{since} 13^23=507<512=2^9\
        13^{4} quad ?< quad 2^{15}qquad \28561 quad < quad 32678
        $$



        Finally, $f(x)$ is decreasing means that if $a>b$, then $f(a) < f(b) < 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 31 at 19:50









        Vasily MitchVasily Mitch

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