Understanding Ln Graph: Form & Deduction of P and L in V = Pe-LQ

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    Graph Ln
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Homework Help Overview

The discussion revolves around the equation V = Pe-LQ, focusing on understanding the graph of lnV against Q and how to deduce the constants P and L from it. Participants are exploring the relationship between the variables and the implications of logarithmic transformations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss taking the natural logarithm of both sides of the equation to simplify the expression. There are questions about applying logarithmic rules correctly and concerns about getting confused in the process.

Discussion Status

Some participants have attempted to apply logarithmic properties but express frustration with their results. Guidance has been offered regarding the manipulation of the logarithmic terms, suggesting that the equation can be transformed into a linear form. There is an ongoing exploration of the correct approach without a clear consensus on the next steps.

Contextual Notes

Participants are encouraged to share their work to identify specific misunderstandings. There is an emphasis on the need to clarify the application of logarithmic rules in the context of the problem.

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If V = Pe-LQ
Where P and L are constants,

Describe the form that a graph of lnV against Q should take and explain how P and L can be deduced?

Very very stuck, please help!
 
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If you take logs of both sides

Ln(V) = Ln(Pe^-LQ)

you should then be able to apply the rules for dealing with logs to the equation and go from there.
 
MalachiK said:
If you take logs of both sides

Ln(V) = Ln(Pe^-LQ)

you should then be able to apply the rules for dealing with logs to the equation and go from there.

I have tried this and just got myself in a massive mess haha, any help?
 
You should post what you have done so that we can see where the problem is. the left hand side is just Ln(V), nothing has to happen to that. On the right you have two terms multiplied by each other.

You know that in general ln(AB) = ln(A) + ln(B), apply this and you should get something that looks like the equation of a straight line... remember y = mx + c?
 

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