Behavior of a linear regression when using a LogLog-Plot












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I am plotting some data on a Log-Log-Plot and wanted to do a linear regression. But because of the Log-Log-Scale it isn't a straight line. It is a line which makes a kink in the center. My supervisor told me that the right straight part represents the constant part of the linear function and the left straight part represents the slope.



Assuming that this is right, I wonder why this is the case. Doing the Logarithmic Math doesn't lead me to a satisfying result:



$$y = mx + c$$
$$Y = log_{10}(mx + c)$$



I have to admit that I am currently not able to simplify or resolve this further to prove and understand the behavior.



I hope someone can help me with this.



Thank you in advance.






real-analysis logarithms graphing-functions







share|cite|improve this question









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    0












    $begingroup$


    I am plotting some data on a Log-Log-Plot and wanted to do a linear regression. But because of the Log-Log-Scale it isn't a straight line. It is a line which makes a kink in the center. My supervisor told me that the right straight part represents the constant part of the linear function and the left straight part represents the slope.



    Assuming that this is right, I wonder why this is the case. Doing the Logarithmic Math doesn't lead me to a satisfying result:



    $$y = mx + c$$
    $$Y = log_{10}(mx + c)$$



    I have to admit that I am currently not able to simplify or resolve this further to prove and understand the behavior.



    I hope someone can help me with this.



    Thank you in advance.






    real-analysis logarithms graphing-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am plotting some data on a Log-Log-Plot and wanted to do a linear regression. But because of the Log-Log-Scale it isn't a straight line. It is a line which makes a kink in the center. My supervisor told me that the right straight part represents the constant part of the linear function and the left straight part represents the slope.



      Assuming that this is right, I wonder why this is the case. Doing the Logarithmic Math doesn't lead me to a satisfying result:



      $$y = mx + c$$
      $$Y = log_{10}(mx + c)$$



      I have to admit that I am currently not able to simplify or resolve this further to prove and understand the behavior.



      I hope someone can help me with this.



      Thank you in advance.






      real-analysis logarithms graphing-functions







      share|cite|improve this question









      $endgroup$




      I am plotting some data on a Log-Log-Plot and wanted to do a linear regression. But because of the Log-Log-Scale it isn't a straight line. It is a line which makes a kink in the center. My supervisor told me that the right straight part represents the constant part of the linear function and the left straight part represents the slope.



      Assuming that this is right, I wonder why this is the case. Doing the Logarithmic Math doesn't lead me to a satisfying result:



      $$y = mx + c$$
      $$Y = log_{10}(mx + c)$$



      I have to admit that I am currently not able to simplify or resolve this further to prove and understand the behavior.



      I hope someone can help me with this.



      Thank you in advance.






      real-analysis logarithms graphing-functions




      real-analysis logarithms graphing-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 8 at 10:04









      birdfreeyahoobirdfreeyahoo

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          $begingroup$

          In fact, you have the change of variables :
          $$begin{cases}
          X=log_{10}(x) quad;quad x=10^X \
          Y=log_{10}(y) quad;quad y=10^Y
          end{cases}$$
          In the new system of axes $(X,Y)$ the equation becomes :
          $$Y=log_{10}(m:10^X+c)$$



          FIRST : For $xto 0$



          $Xto -inftyquad;quad m10^Xto 0 quad;quad Ytolog_{10}(c)$



          $$text{Horizontal asymptote :}quad Y=log_{10}(c)quadtext{in blue on the figure.}$$



          Thus this straight line is related to the constant part of the linear function.



          SECOND : For $xto +infty quad;quad Xto+infty$



          $Y=log_{10}(m:10^X+c)=log_{10}(m:10^X)+log_{10}(1+frac{c}{m:10^X})$



          $Y=log_{10}(m)+X+log_{10}(1+frac{c}{m:10^X})$



          $frac{c}{m:10^X}to 0 quad;quad Ysimeq X+log_{10}(m)$
          $$text{Inclined asymptote}quad Y= X+log_{10}(m)quadtext{in green on the figure}$$
          Thus this straight line is related to the slope of the linear function.



          enter image description here






          share|cite|improve this answer









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            $begingroup$

            In fact, you have the change of variables :
            $$begin{cases}
            X=log_{10}(x) quad;quad x=10^X \
            Y=log_{10}(y) quad;quad y=10^Y
            end{cases}$$
            In the new system of axes $(X,Y)$ the equation becomes :
            $$Y=log_{10}(m:10^X+c)$$



            FIRST : For $xto 0$



            $Xto -inftyquad;quad m10^Xto 0 quad;quad Ytolog_{10}(c)$



            $$text{Horizontal asymptote :}quad Y=log_{10}(c)quadtext{in blue on the figure.}$$



            Thus this straight line is related to the constant part of the linear function.



            SECOND : For $xto +infty quad;quad Xto+infty$



            $Y=log_{10}(m:10^X+c)=log_{10}(m:10^X)+log_{10}(1+frac{c}{m:10^X})$



            $Y=log_{10}(m)+X+log_{10}(1+frac{c}{m:10^X})$



            $frac{c}{m:10^X}to 0 quad;quad Ysimeq X+log_{10}(m)$
            $$text{Inclined asymptote}quad Y= X+log_{10}(m)quadtext{in green on the figure}$$
            Thus this straight line is related to the slope of the linear function.



            enter image description here






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              In fact, you have the change of variables :
              $$begin{cases}
              X=log_{10}(x) quad;quad x=10^X \
              Y=log_{10}(y) quad;quad y=10^Y
              end{cases}$$
              In the new system of axes $(X,Y)$ the equation becomes :
              $$Y=log_{10}(m:10^X+c)$$



              FIRST : For $xto 0$



              $Xto -inftyquad;quad m10^Xto 0 quad;quad Ytolog_{10}(c)$



              $$text{Horizontal asymptote :}quad Y=log_{10}(c)quadtext{in blue on the figure.}$$



              Thus this straight line is related to the constant part of the linear function.



              SECOND : For $xto +infty quad;quad Xto+infty$



              $Y=log_{10}(m:10^X+c)=log_{10}(m:10^X)+log_{10}(1+frac{c}{m:10^X})$



              $Y=log_{10}(m)+X+log_{10}(1+frac{c}{m:10^X})$



              $frac{c}{m:10^X}to 0 quad;quad Ysimeq X+log_{10}(m)$
              $$text{Inclined asymptote}quad Y= X+log_{10}(m)quadtext{in green on the figure}$$
              Thus this straight line is related to the slope of the linear function.



              enter image description here






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                In fact, you have the change of variables :
                $$begin{cases}
                X=log_{10}(x) quad;quad x=10^X \
                Y=log_{10}(y) quad;quad y=10^Y
                end{cases}$$
                In the new system of axes $(X,Y)$ the equation becomes :
                $$Y=log_{10}(m:10^X+c)$$



                FIRST : For $xto 0$



                $Xto -inftyquad;quad m10^Xto 0 quad;quad Ytolog_{10}(c)$



                $$text{Horizontal asymptote :}quad Y=log_{10}(c)quadtext{in blue on the figure.}$$



                Thus this straight line is related to the constant part of the linear function.



                SECOND : For $xto +infty quad;quad Xto+infty$



                $Y=log_{10}(m:10^X+c)=log_{10}(m:10^X)+log_{10}(1+frac{c}{m:10^X})$



                $Y=log_{10}(m)+X+log_{10}(1+frac{c}{m:10^X})$



                $frac{c}{m:10^X}to 0 quad;quad Ysimeq X+log_{10}(m)$
                $$text{Inclined asymptote}quad Y= X+log_{10}(m)quadtext{in green on the figure}$$
                Thus this straight line is related to the slope of the linear function.



                enter image description here






                share|cite|improve this answer









                $endgroup$



                In fact, you have the change of variables :
                $$begin{cases}
                X=log_{10}(x) quad;quad x=10^X \
                Y=log_{10}(y) quad;quad y=10^Y
                end{cases}$$
                In the new system of axes $(X,Y)$ the equation becomes :
                $$Y=log_{10}(m:10^X+c)$$



                FIRST : For $xto 0$



                $Xto -inftyquad;quad m10^Xto 0 quad;quad Ytolog_{10}(c)$



                $$text{Horizontal asymptote :}quad Y=log_{10}(c)quadtext{in blue on the figure.}$$



                Thus this straight line is related to the constant part of the linear function.



                SECOND : For $xto +infty quad;quad Xto+infty$



                $Y=log_{10}(m:10^X+c)=log_{10}(m:10^X)+log_{10}(1+frac{c}{m:10^X})$



                $Y=log_{10}(m)+X+log_{10}(1+frac{c}{m:10^X})$



                $frac{c}{m:10^X}to 0 quad;quad Ysimeq X+log_{10}(m)$
                $$text{Inclined asymptote}quad Y= X+log_{10}(m)quadtext{in green on the figure}$$
                Thus this straight line is related to the slope of the linear function.



                enter image description here







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 6:04









                JJacquelinJJacquelin

                45.7k21858




                45.7k21858






























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