# Exponential models S1 question

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#1
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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2 years ago
#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.
Thoughts??
0
2 years ago
#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.
Perform a direct substitution and rearrange to one of the given forms.
1
#4
(Original post by RDKGames)
Thoughts??
Not a clue if I'm honest.
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2 years ago
#5
(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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#6
(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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2 years ago
#7
(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

What's the last equality?
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#8
(Original post by RDKGames)
Yeah, now notice that

What's the last equality?
x=10^0.4? I'm really confused.
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2 years ago
#9
(Original post by dont know it)
x=10^0.4? I'm really confused.
So you are.

Notice: .

Hence .

So what sort of form is this? or ?
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#10
Why does that first line give x^0.4?
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2 years ago
#11
(Original post by dont know it)
Why does that first line give x^0.4?
Because

Log and exponential are inverse operations of each other. They cancel out.
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#12
(Original post by RDKGames)
Because

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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2 years ago
#13
(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

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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#14
(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 .
Oh right I see thank you. I get it now, at last.
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