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Data are coded using y=Log y and X=Log x to give a linear relationship. The equation of the regression line for the coded data is Y=1.2 + 0.4X
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
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#2
(Original post by dont know it)
Data are coded using y=Log y and X=Log x to give a linear relationship. The equation of the regression line for the coded data is Y=1.2 + 0.4X
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
Data are coded using y=Log y and X=Log x to give a linear relationship. The equation of the regression line for the coded data is Y=1.2 + 0.4X
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
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#3
(Original post by dont know it)
Data are coded using y=Log y and X=Log x to give a linear relationship. The equation of the regression line for the coded data is Y=1.2 + 0.4X
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
Data are coded using y=Log y and X=Log x to give a linear relationship. The equation of the regression line for the coded data is Y=1.2 + 0.4X
a) State whether the relationship between y and x is in the form ax^n or y=kb^x.
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#5
(Original post by dont know it)
Not a clue if I'm honest.
Not a clue if I'm honest.
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(Original post by RDKGames)
Sub in Y = log(y) and X = log(x) and reareange for y. What form do you get??
Sub in Y = log(y) and X = log(x) and reareange for y. What form do you get??
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#7
(Original post by dont know it)
I still don't understand. I rearrange to get y=10^1.2 . 10^0.4logx
I still don't understand. I rearrange to get y=10^1.2 . 10^0.4logx
What's the last equality?
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#9
(Original post by dont know it)
x=10^0.4? I'm really confused.
x=10^0.4? I'm really confused.
Notice:
.Hence
.So what sort of form is this?
or 
?
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#11
(Original post by dont know it)
Why does that first line give x^0.4?
Why does that first line give x^0.4?
Log and exponential are inverse operations of each other. They cancel out.
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(Original post by RDKGames)
Because
Log and exponential are inverse operations of each other. They cancel out.
Because
Log and exponential are inverse operations of each other. They cancel out.
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#13
(Original post by dont know it)
Could you just use the power rule i.e. logx.10 = log10x = x? Or is that wrong?
Could you just use the power rule i.e. logx.10 = log10x = x? Or is that wrong?
But to show that
we just denote 
, take logs to get 
then this is only possible if 
, hence we have that 
.
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(Original post by RDKGames)
Well by the power rule we have
But to show that we just denote , take logs to get then this is only possible if , hence we have that .
Well by the power rule we have
But to show that
we just denote , take logs to get then this is only possible if , hence we have that .
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