https://www.quora.com/profile/Bravewarrior/p-155258163

Here is the question with its solution. I am stuck on part d. I don't know how they got their values for a and b? Some help would be greatly appreciated!

Here is the question with its solution. I am stuck on part d. I don't know how they got their values for a and b? Some help would be greatly appreciated!

(edited 3 months ago)

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/p-155258163

Here is the question with its solution. I am stuck on part d. I don't know how they got their values for a and b? And is it in the form y=ax^n or y=kb^x? Some help would be greatly appreciated!

Here is the question with its solution. I am stuck on part d. I don't know how they got their values for a and b? And is it in the form y=ax^n or y=kb^x? Some help would be greatly appreciated!

P = ab^T

so when you take logs of the model and data, you should be able to relate a and b (or log(a) and log(b)) to the coefficients of a standard linear model between T and log(P)?

Original post by mqb2766

Presuming the model is of the form

P = ab^T

so when you take logs of the model and data, you should be able to relate a and b (or log(a) and log(b)) to the coefficients of a standard linear model between T and log(P)?

P = ab^T

so when you take logs of the model and data, you should be able to relate a and b (or log(a) and log(b)) to the coefficients of a standard linear model between T and log(P)?

(edited 3 months ago)

Original post by pigeonwarrior

After applying logs to both sides of P=ab^T I got logP= loga + Tlogb. In terms of comparing coefficients, I guess logb is the gradient and loga is the y-intercept? After assuming the linear model is logP= mT + C?

Original post by mqb2766

Thats right.

Thank you so much!!! 🙂

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