dont know it
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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.
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RDKGames
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(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.
Thoughts??
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FoxNuke
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(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.
Perform a direct substitution and rearrange to one of the given forms.
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dont know it
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(Original post by RDKGames)
Thoughts??
Not a clue if I'm honest.
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(Original post by dont know it)
Not a clue if I'm honest.
Sub in Y = log(y) and X = log(x) and reareange for y. What form do you get??
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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??
I still don't understand. I rearrange to get y=10^1.2 . 10^0.4logx
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RDKGames
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(Original post by dont know it)
I still don't understand. I rearrange to get y=10^1.2 . 10^0.4logx
Yeah, now notice that 10^{0.4\log x} = 10^{\log x^{0.4}} = ??

What's the last equality?
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(Original post by RDKGames)
Yeah, now notice that 10^{0.4\log x} = 10^{\log x^{0.4}} = ??

What's the last equality?
x=10^0.4? I'm really confused.
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(Original post by dont know it)
x=10^0.4? I'm really confused.
So you are.

Notice: 10^{0.4\log x} = 10^{\log x^{0.4}} = x^{0.4}.

Hence y = 10^{1.2} x^{0.4}.

So what sort of form is this? ax^n or kb^x ?
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Why does that first line give x^0.4?
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RDKGames
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(Original post by dont know it)
Why does that first line give x^0.4?
Because 10^{\log x} = \log(10^x) = x

Log and exponential are inverse operations of each other. They cancel out.
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(Original post by RDKGames)
Because 10^{\log x} = \log(10^x) = x

Log and exponential are inverse operations of each other. They cancel out.
Could you just use the power rule i.e. logx.10 = log10x = x? Or is that wrong?
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(Original post by dont know it)
Could you just use the power rule i.e. logx.10 = log10x = x? Or is that wrong?
Well by the power rule we have \log(10^x) =x \log(10) = x

But to show that 10^{\log x} =x we just denote y = 10^{\log{x}}, take logs to get \log y = \log(10^{\log{x}}) = \log x then this is only possible if x=y, hence we have that x = 10^{\log{x}}.
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(Original post by RDKGames)
Well by the power rule we have \log(10^x) =x \log(10) = x

But to show that 10^{\log x} =x we just denote y = 10^{\log{x}}, take logs to get \log y = \log(10^{\log{x}}) = \log x then this is only possible if x=y, hence we have that x = 10^{\log{x}}.
Oh right I see thank you. I get it now, at last.
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