Why is the Ratio of ln(x) and log(x) a constant?












2












$begingroup$


I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:



$$c^{log(x)}$$ and $$x^{log(c)}$$



grow at the same rate. When figuring this out I ended up with the expression:



$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$



This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?










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  • $begingroup$
    Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
    $endgroup$
    – adfriedman
    2 hours ago


















2












$begingroup$


I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:



$$c^{log(x)}$$ and $$x^{log(c)}$$



grow at the same rate. When figuring this out I ended up with the expression:



$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$



This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
    $endgroup$
    – adfriedman
    2 hours ago
















2












2








2





$begingroup$


I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:



$$c^{log(x)}$$ and $$x^{log(c)}$$



grow at the same rate. When figuring this out I ended up with the expression:



$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$



This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?










share|cite|improve this question









$endgroup$




I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:



$$c^{log(x)}$$ and $$x^{log(c)}$$



grow at the same rate. When figuring this out I ended up with the expression:



$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$



This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?







logarithms






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asked 3 hours ago









user3776749user3776749

286210




286210












  • $begingroup$
    Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
    $endgroup$
    – adfriedman
    2 hours ago




















  • $begingroup$
    Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
    $endgroup$
    – adfriedman
    2 hours ago


















$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
2 hours ago






$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
2 hours ago












5 Answers
5






active

oldest

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2












$begingroup$

For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    If $ln x = a_x$ and $log_{10} x = b_x$ then



    $e^{a_x} = x$ and $10^{b_x} = x$.



    Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.



    So if $10^{k_x} =e^{k_xln 10} = x$ then......



    By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......



    $frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.



    It's just a conversion constant and shouldn't surprise us.



    This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $log_a b$ = $log_c b over log_c a$ is a general rule.
        Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
          $endgroup$
          – Rhys Hughes
          2 hours ago












        • $begingroup$
          Thanks. I edited to correct
          $endgroup$
          – J. W. Tanner
          2 hours ago



















        0












        $begingroup$

        Some of the answers already provided get close to a full explanation, but not quite.



        Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.



        So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$



        Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.






        share|cite|improve this answer









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          5 Answers
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          active

          oldest

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          5 Answers
          5






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          active

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          active

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          2












          $begingroup$

          For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$






              share|cite|improve this answer











              $endgroup$



              For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Michael RozenbergMichael Rozenberg

              99.5k1590189




              99.5k1590189























                  1












                  $begingroup$

                  If $ln x = a_x$ and $log_{10} x = b_x$ then



                  $e^{a_x} = x$ and $10^{b_x} = x$.



                  Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.



                  So if $10^{k_x} =e^{k_xln 10} = x$ then......



                  By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......



                  $frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.



                  It's just a conversion constant and shouldn't surprise us.



                  This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    If $ln x = a_x$ and $log_{10} x = b_x$ then



                    $e^{a_x} = x$ and $10^{b_x} = x$.



                    Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.



                    So if $10^{k_x} =e^{k_xln 10} = x$ then......



                    By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......



                    $frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.



                    It's just a conversion constant and shouldn't surprise us.



                    This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      If $ln x = a_x$ and $log_{10} x = b_x$ then



                      $e^{a_x} = x$ and $10^{b_x} = x$.



                      Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.



                      So if $10^{k_x} =e^{k_xln 10} = x$ then......



                      By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......



                      $frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.



                      It's just a conversion constant and shouldn't surprise us.



                      This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).






                      share|cite|improve this answer









                      $endgroup$



                      If $ln x = a_x$ and $log_{10} x = b_x$ then



                      $e^{a_x} = x$ and $10^{b_x} = x$.



                      Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.



                      So if $10^{k_x} =e^{k_xln 10} = x$ then......



                      By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......



                      $frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.



                      It's just a conversion constant and shouldn't surprise us.



                      This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      fleabloodfleablood

                      69.4k22685




                      69.4k22685























                          0












                          $begingroup$

                          Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?






                              share|cite|improve this answer









                              $endgroup$



                              Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              Chris CulterChris Culter

                              20.4k43584




                              20.4k43584























                                  0












                                  $begingroup$

                                  $log_a b$ = $log_c b over log_c a$ is a general rule.
                                  Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                    $endgroup$
                                    – Rhys Hughes
                                    2 hours ago












                                  • $begingroup$
                                    Thanks. I edited to correct
                                    $endgroup$
                                    – J. W. Tanner
                                    2 hours ago
















                                  0












                                  $begingroup$

                                  $log_a b$ = $log_c b over log_c a$ is a general rule.
                                  Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                    $endgroup$
                                    – Rhys Hughes
                                    2 hours ago












                                  • $begingroup$
                                    Thanks. I edited to correct
                                    $endgroup$
                                    – J. W. Tanner
                                    2 hours ago














                                  0












                                  0








                                  0





                                  $begingroup$

                                  $log_a b$ = $log_c b over log_c a$ is a general rule.
                                  Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.






                                  share|cite|improve this answer











                                  $endgroup$



                                  $log_a b$ = $log_c b over log_c a$ is a general rule.
                                  Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited 2 hours ago

























                                  answered 2 hours ago









                                  J. W. TannerJ. W. Tanner

                                  56611




                                  56611












                                  • $begingroup$
                                    Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                    $endgroup$
                                    – Rhys Hughes
                                    2 hours ago












                                  • $begingroup$
                                    Thanks. I edited to correct
                                    $endgroup$
                                    – J. W. Tanner
                                    2 hours ago


















                                  • $begingroup$
                                    Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                    $endgroup$
                                    – Rhys Hughes
                                    2 hours ago












                                  • $begingroup$
                                    Thanks. I edited to correct
                                    $endgroup$
                                    – J. W. Tanner
                                    2 hours ago
















                                  $begingroup$
                                  Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                  $endgroup$
                                  – Rhys Hughes
                                  2 hours ago






                                  $begingroup$
                                  Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
                                  $endgroup$
                                  – Rhys Hughes
                                  2 hours ago














                                  $begingroup$
                                  Thanks. I edited to correct
                                  $endgroup$
                                  – J. W. Tanner
                                  2 hours ago




                                  $begingroup$
                                  Thanks. I edited to correct
                                  $endgroup$
                                  – J. W. Tanner
                                  2 hours ago











                                  0












                                  $begingroup$

                                  Some of the answers already provided get close to a full explanation, but not quite.



                                  Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.



                                  So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$



                                  Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Some of the answers already provided get close to a full explanation, but not quite.



                                    Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.



                                    So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$



                                    Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Some of the answers already provided get close to a full explanation, but not quite.



                                      Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.



                                      So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$



                                      Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Some of the answers already provided get close to a full explanation, but not quite.



                                      Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.



                                      So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$



                                      Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 1 hour ago









                                      heropupheropup

                                      63k66199




                                      63k66199






























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