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

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The discussion focuses on understanding the graph of ln(V) against Q based on the equation V = Pe^-LQ. By taking the natural logarithm of both sides, the equation simplifies to ln(V) = ln(P) - LQ, which resembles the equation of a straight line. Participants emphasize applying logarithmic rules, particularly ln(AB) = ln(A) + ln(B), to separate terms effectively. The constants P and L can be deduced from the slope and intercept of the resulting linear graph. Clarification on the steps taken is encouraged to resolve any confusion in the calculations.
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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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