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

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SUMMARY

The discussion centers on the mathematical expression V = Pe^-LQ and its transformation into a linear graph by taking the natural logarithm. By applying logarithmic properties, specifically ln(AB) = ln(A) + ln(B), the equation can be rewritten as ln(V) = ln(P) - LQ. This results in a linear relationship where ln(V) is plotted against Q, allowing for the deduction of constants P and L from the slope and intercept of the graph. Participants emphasize the importance of correctly applying logarithmic rules to avoid confusion.

PREREQUISITES
  • Understanding of logarithmic properties, particularly ln(AB) = ln(A) + ln(B)
  • Familiarity with linear equations in the form y = mx + c
  • Basic knowledge of constants in mathematical equations
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the properties of logarithms in depth, focusing on their applications in equations
  • Learn how to derive linear equations from exponential functions
  • Explore graphing techniques for linear relationships
  • Investigate methods for determining slope and intercept from plotted data
USEFUL FOR

Students and professionals in mathematics, physics, or engineering who are working with exponential functions and their graphical representations. This discussion is particularly beneficial for those learning about logarithmic transformations and linear modeling.

georgiemuc
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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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