Prove that two functions grow at equal rate












0












$begingroup$


Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.



I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?



I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.






logarithms







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$endgroup$












  • $begingroup$
    I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
    $endgroup$
    – user376343
    Nov 29 '18 at 15:36












  • $begingroup$
    You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 15:45
















0












$begingroup$


Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.



I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?



I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.






logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
    $endgroup$
    – user376343
    Nov 29 '18 at 15:36












  • $begingroup$
    You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 15:45














0












0








0





$begingroup$


Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.



I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?



I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.






logarithms







share|cite|improve this question











$endgroup$




Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.



I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?



I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.






logarithms




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 15:33









user376343

3,3282825




3,3282825










asked Nov 29 '18 at 15:32









mathisharddmathishardd

1




1












  • $begingroup$
    I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
    $endgroup$
    – user376343
    Nov 29 '18 at 15:36












  • $begingroup$
    You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 15:45


















  • $begingroup$
    I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
    $endgroup$
    – user376343
    Nov 29 '18 at 15:36












  • $begingroup$
    You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
    $endgroup$
    – Ross Millikan
    Nov 29 '18 at 15:45
















$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36






$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36














$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45




$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45










2 Answers
2






active

oldest

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0












$begingroup$

Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
      $endgroup$
      – mathishardd
      Nov 29 '18 at 16:00






    • 1




      $begingroup$
      Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
      $endgroup$
      – Frobenius
      Nov 29 '18 at 16:05













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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
    Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
      Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
        Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.






        share|cite|improve this answer









        $endgroup$



        Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
        Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 15:47









        Ross MillikanRoss Millikan

        293k23197371




        293k23197371























            0












            $begingroup$

            Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
              $endgroup$
              – mathishardd
              Nov 29 '18 at 16:00






            • 1




              $begingroup$
              Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
              $endgroup$
              – Frobenius
              Nov 29 '18 at 16:05


















            0












            $begingroup$

            Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
              $endgroup$
              – mathishardd
              Nov 29 '18 at 16:00






            • 1




              $begingroup$
              Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
              $endgroup$
              – Frobenius
              Nov 29 '18 at 16:05
















            0












            0








            0





            $begingroup$

            Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...






            share|cite|improve this answer









            $endgroup$



            Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 '18 at 15:54









            FrobeniusFrobenius

            613




            613












            • $begingroup$
              Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
              $endgroup$
              – mathishardd
              Nov 29 '18 at 16:00






            • 1




              $begingroup$
              Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
              $endgroup$
              – Frobenius
              Nov 29 '18 at 16:05




















            • $begingroup$
              Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
              $endgroup$
              – mathishardd
              Nov 29 '18 at 16:00






            • 1




              $begingroup$
              Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
              $endgroup$
              – Frobenius
              Nov 29 '18 at 16:05


















            $begingroup$
            Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
            $endgroup$
            – mathishardd
            Nov 29 '18 at 16:00




            $begingroup$
            Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
            $endgroup$
            – mathishardd
            Nov 29 '18 at 16:00




            1




            1




            $begingroup$
            Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
            $endgroup$
            – Frobenius
            Nov 29 '18 at 16:05






            $begingroup$
            Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
            $endgroup$
            – Frobenius
            Nov 29 '18 at 16:05




















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