Behavior of a linear regression when using a LogLog-Plot
$begingroup$
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
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
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add a comment |
$begingroup$
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
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
$endgroup$
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
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
asked Jan 8 at 10:04
birdfreeyahoobirdfreeyahoo
1032
1032
add a comment |
add a comment |
1 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.
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
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add a comment |
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1 Answer
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1 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.
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
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add a comment |
$begingroup$
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.
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
$endgroup$
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.
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
answered Jan 9 at 6:04
JJacquelinJJacquelin
45.7k21858
45.7k21858
add a comment |
add a comment |
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