Elasticity and logarithms












3












$begingroup$


Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$



Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:



$$ eta = b = frac{ d log y}{d log x} $$



Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?



Here's a snapshot from the book:



enter image description here






elasticity







share|improve this question









$endgroup$








  • 1




    $begingroup$
    About half of your questions have been answered, consider accepting some.
    $endgroup$
    – denesp
    Jan 13 at 13:24






  • 1




    $begingroup$
    Thank you for pointing that out. I'll do so right away.
    $endgroup$
    – WorldGov
    Jan 13 at 13:44
















3












$begingroup$


Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$



Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:



$$ eta = b = frac{ d log y}{d log x} $$



Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?



Here's a snapshot from the book:



enter image description here






elasticity







share|improve this question









$endgroup$








  • 1




    $begingroup$
    About half of your questions have been answered, consider accepting some.
    $endgroup$
    – denesp
    Jan 13 at 13:24






  • 1




    $begingroup$
    Thank you for pointing that out. I'll do so right away.
    $endgroup$
    – WorldGov
    Jan 13 at 13:44














3












3








3





$begingroup$


Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$



Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:



$$ eta = b = frac{ d log y}{d log x} $$



Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?



Here's a snapshot from the book:



enter image description here






elasticity







share|improve this question









$endgroup$




Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$



Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:



$$ eta = b = frac{ d log y}{d log x} $$



Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?



Here's a snapshot from the book:



enter image description here






elasticity




elasticity






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 13 at 13:09









WorldGovWorldGov

393114




393114








  • 1




    $begingroup$
    About half of your questions have been answered, consider accepting some.
    $endgroup$
    – denesp
    Jan 13 at 13:24






  • 1




    $begingroup$
    Thank you for pointing that out. I'll do so right away.
    $endgroup$
    – WorldGov
    Jan 13 at 13:44














  • 1




    $begingroup$
    About half of your questions have been answered, consider accepting some.
    $endgroup$
    – denesp
    Jan 13 at 13:24






  • 1




    $begingroup$
    Thank you for pointing that out. I'll do so right away.
    $endgroup$
    – WorldGov
    Jan 13 at 13:44








1




1




$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24




$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24




1




1




$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44




$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44










2 Answers
2






active

oldest

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4












$begingroup$

Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$






share|improve this answer









$endgroup$





















    4












    $begingroup$

    Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
    $$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
    Rearranging:
    $$frac{dy/y}{dx/x}=eta=b$$






    share|improve this answer











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      4












      $begingroup$

      Because $a$ is a parameter, and so
      $$
      eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
      $$






      share|improve this answer









      $endgroup$


















        4












        $begingroup$

        Because $a$ is a parameter, and so
        $$
        eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
        $$






        share|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Because $a$ is a parameter, and so
          $$
          eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
          $$






          share|improve this answer









          $endgroup$



          Because $a$ is a parameter, and so
          $$
          eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
          $$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 13 at 13:21









          denespdenesp

          12.5k32248




          12.5k32248























              4












              $begingroup$

              Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
              $$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
              Rearranging:
              $$frac{dy/y}{dx/x}=eta=b$$






              share|improve this answer











              $endgroup$


















                4












                $begingroup$

                Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
                $$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
                Rearranging:
                $$frac{dy/y}{dx/x}=eta=b$$






                share|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
                  $$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
                  Rearranging:
                  $$frac{dy/y}{dx/x}=eta=b$$






                  share|improve this answer











                  $endgroup$



                  Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
                  $$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
                  Rearranging:
                  $$frac{dy/y}{dx/x}=eta=b$$







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Jan 13 at 17:57

























                  answered Jan 13 at 14:25









                  Adam BaileyAdam Bailey

                  3,7041026




                  3,7041026






























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